#岩井PC全問

def solve_A():
    N, M = map(int, input().split())
    ans = 0
    for _ in range(N):
        Si, Ri = input().split()
        if ( Ri := int(Ri) ) >= 1200 and any( Si[p] == 'x' for p in range(4) ):
            ans += 1
    print(ans)


def solve_B():
    M = 30
    x = 5 * 10 ** 8
    for _ in range(M):
        print(x)
        if ( r := int(input()) ) == 1:
            return
        else:
            x = - ( - x // 2 )
    

def solve_C():
    X, Y = map(int, input().split())
    P = [i + 1 for i in range(X * Y)]
    for i in range(X - 1, X * Y, X):
        P[i] = (i + X) % (X * Y)
    for i in range(X * Y):
        print(i + 1, P[i] + 1)


def solve_D():
    N, M, P, Y = map(int, input().split())
    G = [[] for _ in range(N)]
    for _ in range(M):
        Ai, Bi, Ci = map(int, input().split())
        G[Ai - 1].append((Bi - 1, Ci))
        G[Bi - 1].append((Ai - 1, Ci))
    S = [Y + 1] * N
    for _ in range(P):
        Di, Ei = map(int, input().split())
        assert S[Di - 1] == Y + 1
        S[Di - 1] = Ei

    #街0から各街に最短で移動したときのコストを計算
    import heapq
    Q = [(0, 0)]
    D = [Y + 1] * N
    D[0] = 0
    while Q:
        cost, now = heapq.heappop(Q)
        if D[now] < cost:
            continue
        D[now] = cost
        for nxt, d in G[now]:
            if D[nxt] > D[now] + d:
                D[nxt] = D[now] + d
                heapq.heappush(Q, (D[nxt], nxt))

    #各街に最短で着いたら買い占め
    ans = 0
    for now in range(N):
        left = Y - D[now]
        price = S[now]
        if left >= price:
            ans = max(ans, left // price)
    print(ans)


def solve_E():
    N, D, K = map(int, input().split())
    A = list(map(int, input().split()))
    C = list(map(int, input().split()))

    #DP[i := 考慮した問題数][c := 美しさ][d := 採用数] = 最大の満足度 の遷移をひとつのDPで
    DP = [[- 10 ** 18] * (D + 1) for _ in range(K + 1)]
    DP[0][0] = 0
    for Ai, Ci in zip(A, C):  #採用の遷移を逆順ループで
        for d in range(D - 1, -1, -1):
            for c in range(K, -1, -1):
                if DP[c][d] == - 10 ** 18:
                    continue
                nd = d + 1
                nc = min(c + Ci, K)
                DP[nc][nd] = max(DP[nc][nd], DP[c][d] + Ai)
    print( DP[c][d] if DP[c := K][d := D] != - 10 ** 18 else 'No')


def solve_F():
    S = input()
    N = len(S)
    #0を2個飛ばしで左に寄せられる。ただし1が奇数の場合、操作不可能
    #1110 → 1011 ･･･ ■ となる ので、飛ばせる1の個数を持ってシミュレートすればよさそう
    RLE = []
    Lt = 0
    for i in range(N):
        if i == N - 1 or S[i] != S[i + 1]:
            RLE.append((S[i], i - Lt + 1))
            Lt = i + 1
    n = len(RLE)
    ans = 0
    cont_1 = 0
    for Si, c in RLE:
        if Si == '0':
            cont_1 >>= 1
            cont_1 <<= 1
            ans += c * ( cont_1 >> 1 )
        if Si == '1':
            cont_1 += c
    print(ans)
    


def solve_G():
    N = int(input())
    A = list(map(int, input().split()))
    B = list(map(int, input().split()))

    #自信はないがソート順マッチでいいんじゃないか？
    A.sort()
    B.sort()

    #C[i]: 左詰マッチしたときの累積不満度
    #D[i]: 右詰マッチしたときの累積不満度
    C = [0] * (N + 1)
    for i in range(N - 1):
        C[i + 1] = C[i] + abs(A[i] - B[i])
    C[N] = C[N - 1]
    D = [0] * (N + 1)
    for i in range(N - 1, 0, -1):
        D[i] = D[i + 1] - abs(A[i] - B[i - 1])
    D[0] = D[1]

    min_comp = 10 ** 18
    ans = []
    for i in range(N):
        cnt = ( C[i] - C[0] ) + ( D[N] - D[i + 1] )
        if min_comp > cnt:
            min_comp = cnt
            ans = []
        if min_comp == cnt:  #種類数なので重複を許さない
            if len(ans) > 0 and ans[-1] == A[i]:
                continue
            ans.append( A[i] )
    print( len(ans) )
    print( *ans )


def solve_H():
    N = int(input())
    H = list(map(int, input().split()))

    #stackでシミュレート
    Q = [(0, 10 ** 9 + 2)]
    ans = 0  #現在の水色の量
    for i, Hi in enumerate(H):
        c = (i + 1) & 1
        d = 0  #今回染めた量
        while d < Hi:
            color, dist = Q.pop()  #足から取り出す
            if d + dist > Hi:  #全部やると塗りすぎになる場合
                diff = (d + dist) - Hi
                Q.append((color, diff))
                Q.append((color, dist - diff))
                continue  #一旦仕切り直し
            assert d + dist <= Hi

            #一旦水色を削除
            if color == 1:
                ans -= dist
            d += dist

        #色cに染めて戻す
        if c == 1:
            ans += d
        Q.append((c, d))
        print(ans)
            
            
def solve_I():
    #昔自分でも作ったことあるかも
    H, W = map(int, input().split())
    print('?', 1, 1)
    c1 = int(input())
    if c1 == 0:
        print('!', 1, 1)
        return
    if H > 1:  #場合分け
        if H == 2 and W == 1:
            print('!', 2, 1)
            return
        print('?', 2, 1)
        c2 = int(input())
        if c2 == 0:
            print('!', 2, 1)
            return
        if c1 < c2:
            assert c1 + 1 == c2
            h = 1
        else:
            assert c1 > c2
            diff = c1 - c2
            h = 1 + (diff + 1) // 2
        b = (h - 1) ** 2
        d = c1 - b
        c = max(0, int(d ** 0.5) - 2)
        while c ** 2 < d: c += 1
        w = 1 + c
        print('!', h, w)
        return
    else:
        assert H == 1 and W > 1
        if H == 1 and W == 2:
            print('!', 1, 2)
            return
        print('?', 1, 2)
        c2 = int(input())
        if c2 == 0:
            print('!', 1, 2)
            return
        if c1 < c2:
            assert c1 + 1 == c2
            w = 1
        else:
            assert c1 > c2
            diff = c1 - c2
            w = 1 + (diff + 1) // 2
        print('!', 1, w)
        return
        
        
        
#片っ端から提出
solve_B()