# input
import sys
# input = sys.stdin.readline
II = lambda : int(input())
MI = lambda : map(int, input().split())
LI = lambda : [int(a) for a in input().split()]
SI = lambda : input().rstrip()
LLI = lambda n : [[int(a) for a in input().split()] for _ in range(n)]
LSI = lambda n : [input().rstrip() for _ in range(n)]
MI_1 = lambda : map(lambda x:int(x)-1, input().split())
LI_1 = lambda : [int(a)-1 for a in input().split()]

mod = 998244353
inf = 1001001001001001001
ordalp = lambda s : ord(s)-65 if s.isupper() else ord(s)-97
ordallalp = lambda s : ord(s)-39 if s.isupper() else ord(s)-97
yes = lambda : print("Yes")
no = lambda : print("No")
yn = lambda flag : print("Yes" if flag else "No")

prinf = lambda ans : print(ans if ans < 1000001001001001001 else -1)
alplow = "abcdefghijklmnopqrstuvwxyz"
alpup = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
alpall = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"
URDL = {'U':(-1,0), 'R':(0,1), 'D':(1,0), 'L':(0,-1)}
DIR_4 = [[-1,0],[0,1],[1,0],[0,-1]]
DIR_8 = [[-1,0],[-1,1],[0,1],[1,1],[1,0],[1,-1],[0,-1],[-1,-1]]
DIR_BISHOP = [[-1,1],[1,1],[1,-1],[-1,-1]]
prime60 = [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59]
sys.set_int_max_str_digits(0)
# sys.setrecursionlimit(10**6)
# import pypyjit
# pypyjit.set_param('max_unroll_recursion=-1')

from collections import defaultdict,deque
from heapq import heappop,heappush
from bisect import bisect_left,bisect_right
DD = defaultdict
BSL = bisect_left
BSR = bisect_right


from math import isqrt
from random import randint

def gcd(x, y):
    """ x < y """
    while y:
        x, y = y, x%y
    return x

def is_prime(num):
    """ 1 <= x < 1<<64 """
    if num < 4: return num > 1
    if not num&1: return False
    
    d, s = num-1, 0
    while not d&1:
        d >>= 1
        s += 1
        
    tests = (2,7,61) if num < 4759123141 else (2,325,9375,28178,450775,9780504,1795265022)
        
    for test in tests:
        if test >= num: return True
        t = pow(test, d, num)
        if 1 < t < num-1:
            for _ in range(s-1):
                t = t*t%num
                if t == num-1: break
            else:
                return False
    return True

def find_prime(n):
    b = n.bit_length() - 1
    b = (b >> 2) << 2
    m = (1 << (b >> 3)) << 1
    while True:
        c = randint(1, n - 1)
        y = 0
        g = q = r = 1
        while g == 1:
            x = y
            for _ in range(r):
                y = (y * y + c) % n
            k = 0
            while k < r and g == 1:
                ys = y
                for _ in range(min(m, r - k)):
                    y = (y * y + c) % n
                    q = q * abs(x - y) % n
                g = gcd(q, n)
                k += m
            r <<= 1
        if g == n:
            g = 1
            y = ys
            while g == 1:
                y = (y * y + c) % n
                g = gcd(abs(x - y), n)
        if g == n:
            continue
        if is_prime(g):
            return g
        elif is_prime(n // g):
            return n // g
        else:
            n = g

def _primefactor(n):
    result = []
    for p in range(2, 500):
        if p * p > n:
            break
        c = 0
        while n%p == 0:
            n //= p
            c += 1
        if c:
            result.append(p)
    
    while n > 1 and not is_prime(n):
        p = find_prime(n)
        while n % p == 0:
            n //= p
        result.append(p)
    if n > 1: result.append(n)
    return result

def primefact(n, deduplicate = True):
    if deduplicate == False:
        return _primefactor(n)
    result = dict()
    for p in range(2, 500):
        if p * p > n:
            break
        c = 0
        while n%p == 0:
            n //= p
            c += 1
        if c:
            result[p] = c
    
    while n > 1 and not is_prime(n):
        p = find_prime(n)
        c = 0
        while n % p == 0:
            n //= p
            c += 1
        result[p] = c
    if n > 1: result[n] = 1
    return result

def divisors_naive(n):
    divs_small, divs_big = [], []
    i = 1
    while i*i <= n:
        if n % i == 0:
            divs_small.append(i)
            if i != n//i:
                divs_big.append(n//i)
        i += 1
    return divs_small + divs_big[::-1]

def divisors(n):
    if n == 1: return [1]
    if n <= 100_000_000: # 10 ** 8
        return divisors_naive(n)
    
    pf = primefact(n)
    ps = list(pf.keys())
    es = list(pf.values())
    us = [p ** e for p,e in zip(ps, es)]
    
    l = len(es)
    nes = [0] * (l + 1)
    r = 1
    res = [1]
    while True:
        nes[0] += 1
        for i in range(l):
            if nes[i] > es[i]:
                if i+1 == l:
                    res.sort()
                    return res
                nes[i] = 0
                nes[i+1] += 1
                r //= us[i]
            else:
                r *= ps[i]
                break
        res.append(r)
        
def totient(n):
    """
    totient(n) = #{ m | (m,n) = 1, 1 <= m <= n }
    """
    pf = _primefactor(n)
    for p in pf:
        n //= p
        n *= p - 1
    return n

def mobius(n):
    pf = primefact(n)
    r = 1
    for p,e in pf.items():
        if e >= 2: return 0
        r *= -1
    return r

def primitive_root(p):
    """ p : prime """
    if p == 2: return 1
    
    r = p - 1
    tests = []
    for q in range(2, 500):
        if q * q > r:
            break
        if r % q == 0:
            while r % q == 0:
                r //= q
            tests.append((p - 1) // q)
    
    while r > 1 and not is_prime(r):
        q = find_prime(r)
        while r % q == 0:
            r //= q
        tests.append((p - 1) // q)
    if r > 1: tests.append((p - 1) // r)
    
    res = 2
    while True:
        for test in tests:
            if pow(res, test, p) == 1:
                break
        else:
            return res
        res = randint(3, p - 2)

"""
つまり B 「ある i, j, k が一致するか？」のみ聞ける

n <= 1e6
q = 1600


n ができればできる？
- k を fai(n) にして聞けば ok
- i = 0, j != 0 で 0 も回避可能

ということで n を特定する問題


そもそもおなじ x を期待するのってそんなに簡単なのか？

支配的な項は k とおもうべきかも

乱択かも？

たとえば多くの N で 1 を返す関数を考える
-> 約数がおおいとか？

誕生日攻撃とか？
1000 回でおよそ 4 割成功 -> カス
1600 で 7 割

---

直接 N を返す関数を作るべきだなー
たとえば i にめっちゃ約数が多い数を選ぶみたいな戦法もありうる

---

それは k でやったほうが効率的

i と j が両方あるの何の意味があるんでしょう？
たとえば偶数にできる -> ?

---

BS-GS とかに寄せるべき？

乱択でまともにできる気がせん。

---

適当な集合で x, y in S で (x - y) % N == 0 が存在すればよいという説 

できそう

できません。たすけて

あー、サイズを半分にできるのか
"""
from random import randint

# lim = 10 ** 6
# s = []
# ok = {0,}
# # while len(ok) < lim + 1:
# for _ in range(1400):
#     best = (-1, -1)
#     for _ in range(100):
#         x = randint(1, lim)
#         add = sum(1 for y in s if abs(x - y) not in ok)
#         best = max(best, (add, x))
#     x = best[1]
#     for y in s:
#         ok.add(abs(x - y))
#     s.append(x)
#     # print(len(ok), len(s))

b = 710
s = [*range(0, b)] + [5 * 10 ** 5 + b * i for i in range(b)]

# print(len(s))
# print(s)

t = II()
for i in range(n):
    qc = 0
    def ask(i, j, k):
        global qc
        qc += 1
        print("?", i, j, k)
        return int(input())

    k = 10 ** 6 + 33

    cnt = DD(list)
    for x in s:
        cnt[ask(x, 0, k)].append(x)

    n = 0
    for v, c in cnt.items():
        c.sort()
        for i in range(len(c) - 1):
            n = gcd(n, c[j] - c[i])


    # n を求めたあと
    phi = totient(n)
    phi = max(phi, 2)

    b = DD(int)
    while qc < 1600:
        i = randint(0, 10 ** 9)
        b[ask(i, 0, phi)] += 1

    ans = max((v, k) for k, v in b.items())[1]
    print("!", ans)