def bitcount(x): """ 立っているbitの数 """ x = (x & 0x5555555555555555) + (x >> 1 & 0x5555555555555555) x = (x & 0x3333333333333333) + (x >> 2 & 0x3333333333333333) x = (x & 0x0f0f0f0f0f0f0f0f) + (x >> 4 & 0x0f0f0f0f0f0f0f0f) x = (x & 0x00ff00ff00ff00ff) + (x >> 8 & 0x00ff00ff00ff00ff) x = (x & 0x0000ffff0000ffff) + (x >> 16 & 0x0000ffff0000ffff) return (x & 0x00000000ffffffff) + (x >> 32 & 0x00000000ffffffff) def calc(i, j, s, C): """ :param i: 自分たちの行の知ってる知ってないの状態 :param j: 前の行の挙げてる挙げてないの状態 :param s: 自分たちの行のシャイ度 :param C: 横幅 :return: 自分たちがどのような挙手状態になるか """ ret = 0 p = [0] * C for c in range(C): c1 = 1 << c if i & c1: p[c] = 4 - s[c] if j & c1: p[c] += 1 if p[c] >= 4: ret |= c1 pret = 0 while pret != ret: for c in range(C): c1 = 1 << c if ret & c1 and pret ^ c1: if c > 0: p[c - 1] += 1 if p[c - 1] >= 4: ret |= 1 << (c - 1) if c < C - 1: p[c + 1] += 1 if p[c + 1] >= 4: ret |= 1 << (c + 1) pret = ret # print((bin(i), bin(j), s, bin(ret))) return ret def solve(): R, C = map(int, input().split()) P = [] S = [] input() for r in range(R): P.append(list(map(int, input().split()))) input() for r in range(R): S.append(list(map(int, input().split()))) ans = 0 Y = [[0.0] * (1 << C) for r in range(R)] # Y[r][i]: r 行目の手を挙げる挙げないの組み合わせが i になる確率 for r in range(R): yr = Y[r] for i in range(1 << C): x = 1.0 for c in range(C): if i & (1 << c): x *= P[r][c] / 100 else: x *= (100 - P[r][c]) / 100 # x: 自分たちの知ってる知ってないの組み合わせが i になる確率 for j in range(1 << C): if r > 0: y = Y[r - 1][j] elif j == 0: y = 1.0 else: continue z = calc(i, j, S[r], C) yr[z] += y * x for i in range(1 << C): # print((bin(i), bitcount(i), yr[i])) ans += bitcount(i) * yr[i] print(ans) if __name__ == '__main__': solve()