The answer is a multinomial coefficient \(\dfrac{(C_1+C_2+\ldots+C_N)!}{C_1!C_2!\dots C_N!}\). A multinomial coefficient can be represented as a product of binomial coefficients:

\[\dfrac{(C_1+C_2+\ldots+C_N)!}{C_1!C_2!\dots C_N!}=\binom{C_1}{C_1}\binom{C_1+C_2}{C_2}\dots \binom{C_1+C_2+\dots +C_N}{C_N}\]

There are several ways to compute binomial coefficients modulo \(M\) (which is not necessarily a prime). This time we have the constraint \(\sum C\leq 5000\), so the following formula on binomial coefficients allows us to precalculate \(\binom{n}{k}\ (0\leq k\leq n\leq 5000)\) easily.

\[\binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}\]

Therefore, each test case can be solved in \(O(N)\) time.

T, mod = map(int, input().split())

K = 5010

binom = [[0] * K for i in range(K)]
binom[0][0] = 1

for n in range(1, K):
    binom[n][0] = 1
    for k in range(1, n + 1):
        binom[n][k] = (binom[n - 1][k - 1] + binom[n - 1][k]) % mod

for _ in range(T):
    N = int(input())
    C = list(map(int, input().split()))
    ans = 1
    s = 0
    for i in C:
        s += i
        ans *= binom[s][i]
        ans %= mod
    print(ans % mod)