def isPrimeMR(n): if n==1: return 0 d = n - 1 d = d // (d & -d) L = [2, 3, 5, 7, 11, 13, 17] if n in L: return 1 for a in L: t = d y = pow(a, t, n) if y == 1: continue while y != n - 1: y = (y * y) % n if y == 1 or t == n - 1: return 0 t <<= 1 return 1 def findFactorRho(n): from math import gcd m = 1 << n.bit_length() // 8 for c in range(1, 99): f = lambda x: (x * x + c) % n y, r, q, g = 2, 1, 1, 1 while g == 1: x = y for i in range(r): y = f(y) k = 0 while k < r and g == 1: ys = y for i in range(min(m, r - k)): y = f(y) q = q * abs(x - y) % n g = gcd(q, n) k += m r <<= 1 if g == n: g = 1 while g == 1: ys = f(ys) g = gcd(abs(x - ys), n) if g < n: if isPrimeMR(g): return g elif isPrimeMR(n // g): return n // g return findFactorRho(g) def primeFactor(n): i = 2 ret = {} rhoFlg = 0 while i*i <= n: k = 0 while n % i == 0: n //= i k += 1 if k: ret[i] = k i += 1 + i % 2 if i == 101 and n >= 2 ** 20: while n > 1: if isPrimeMR(n): ret[n], n = 1, 1 else: rhoFlg = 1 j = findFactorRho(n) k = 0 while n % j == 0: n //= j k += 1 ret[j] = k if n > 1: ret[n] = 1 if rhoFlg: ret = {x: ret[x] for x in sorted(ret)} return ret import sys from itertools import permutations from heapq import heappop,heappush from collections import deque import random import bisect input = lambda :sys.stdin.readline().rstrip() mi = lambda :map(int,input().split()) li = lambda :list(mi()) K = 17 popcount = [0] * (1<> i & 1: popcount[S] ^= 1 mod = 998244353 T,M = mi() pf = primeFactor(M) """ M_div = [1] for p in pf: e = pf[p] nxt_M_div = [] for a in M_div: t = 1 for i in range(e+1): nxt_M_div.append(a * t) t *= p M_div = nxt_M_div M_div.sort() """ M_prime = [pow(p,pf[p]) for p in pf] M_n = len(M_prime) def solve1(N,B,C,D,A): dic = {} w = B for a in A: if M % a: w = (C * w + D) % mod continue if a not in dic: dic[a] = 1 dic[a] *= w + 1 dic[a] %= mod w = (C * w + D) % mod for p in pf: for d in M_div: if d not in dic: continue if M % (d*p): continue if d*p not in dic: dic[d*p] = 1 dic[d*p] *= dic[d] dic[d*p] %= mod for d in dic: dic[d] = (dic[d] - 1 ) % mod for p in pf: for d in M_div[::-1]: if d % p: continue if d // p not in dic: continue dic[d] -= dic[d//p] dic[d] %= mod if M not in dic: return 0 return dic[M] def solve2(N,B,C,D,A): dp = [1] * (1<>i & 1 == 0: dp[S^t] *= dp[S] dp[S^t] %= mod res = sum(dp) % mod if M_n & 1: res = (mod-res) % mod return res for _ in range(T): N,B,C,D = mi() A = li() if M!=1: print(solve2(N,B,C,D,A)) else: print((solve2(N,B,C,D,A)-1) % mod) #print(solve1(N,B,C,D,A))