'''
转成异或方程组求解，每个解数值为0表示开关没有动过，1表示开关拨动过一次
假设最后自由元数量是k，就有2^k种不同方案
'''


case_num = int(input())
for _ in range(case_num):
    n = int(input())
    start_stat = list(map(int, input().split()))
    end_stat = list(map(int, input().split()))

    m = [[0] * (n+1) for _ in range(n)]
    for i in range(n):
        m[i][i] = 1
    while True:
        a, b = map(int, input().split())
        if a == 0 and b == 0:
            break
        a, b = a-1, b-1
        m[b][a] = 1

    for i in range(n):
        m[i][n] = start_stat[i] ^ end_stat[i]

    # 高斯消元
    for col in range(0, n):
        max_val, max_row = m[col][col], col
        for row in range(col + 1, n):
            if max_val == 1:
                break

            if m[row][col] > max_val:
                max_val, max_row = m[row][col], row
                break

        if max_val == 0:
            continue

        m[col], m[max_row] = m[max_row], m[col]

        for row in range(col + 1, n):
            if m[row][col] == 0:
                continue

            for j in range(col, n + 1):
                m[row][j] ^= m[col][j]

    flag = 0
    ans = [0] * n
    solution_cnt = 1
    for i in range(n - 1, -1, -1):
        val = m[i][n]
        s = 0
        for j in range(n - 1, i, -1):
            if m[i][j] == 1:
                s ^= ans[j]
        val ^= s

        if val == 0 and m[i][i] == 0:
            # 多组解
            flag = 1
            solution_cnt *= 2

        if val == 1 and m[i][i] == 0:
            # 无解
            flag = 2
            break

        else:
            ans[i] = val

    if flag == 2:
        print("Oh,it's impossible~!!")
    else:
        print(solution_cnt)