This is the easy version of the problem. The only difference between the easy version and the hard version is the constraints on n. You can only make hacks if both versions are solved.

A permutation of 1, 2, …, n is a sequence of n integers, where each integer from 1 to n appears exactly once. For example, [2,3,1,4] is a permutation of 1, 2, 3, 4, but [1,4,2,2] isn't because 2 appears twice in it.

Recall that the number of inversions in a permutation a_1, a_2, …, a_n is the number of pairs of indices (i, j) such that i < j and a_i > a_j.

Let p and q be two permutations of 1, 2, …, n. Find the number of permutation pairs (p,q) that satisfy the following conditions:

  * p is lexicographically smaller than q. 
  * the number of inversions in p is greater than the number of inversions in q. 



Print the number of such pairs modulo mod. Note that mod may not be a prime.

Input

The only line contains two integers n and mod (1≤ n≤ 50, 1≤ mod≤ 10^9).

Output

Print one integer, which is the answer modulo mod.

Example

Input


4 403458273


Output


17

Note

The following are all valid pairs (p,q) when n=4.

  * p=[1,3,4,2], q=[2,1,3,4], 
  * p=[1,4,2,3], q=[2,1,3,4], 
  * p=[1,4,3,2], q=[2,1,3,4], 
  * p=[1,4,3,2], q=[2,1,4,3], 
  * p=[1,4,3,2], q=[2,3,1,4], 
  * p=[1,4,3,2], q=[3,1,2,4], 
  * p=[2,3,4,1], q=[3,1,2,4], 
  * p=[2,4,1,3], q=[3,1,2,4], 
  * p=[2,4,3,1], q=[3,1,2,4], 
  * p=[2,4,3,1], q=[3,1,4,2], 
  * p=[2,4,3,1], q=[3,2,1,4], 
  * p=[2,4,3,1], q=[4,1,2,3], 
  * p=[3,2,4,1], q=[4,1,2,3], 
  * p=[3,4,1,2], q=[4,1,2,3], 
  * p=[3,4,2,1], q=[4,1,2,3], 
  * p=[3,4,2,1], q=[4,1,3,2], 
  * p=[3,4,2,1], q=[4,2,1,3]. 
import sys
input = sys.stdin.readline



def make_tree(n):
    i = 2
    while True:
        if i >= n * 2:
            tree = [0] * i
            break
        else:
            i *= 2
    return tree

def update(i, x):
    i += len(tree) // 2
    tree[i] = x
    i //= 2
    while True:
        if i == 0:
            break
        tree[i] = tree[2 * i] + tree[2 * i + 1]
        i //= 2
    return

def get_sum(s, t):
    s += len(tree) // 2
    t += len(tree) // 2
    ans = 0
    while True:
        if s > t:
            break
        if s % 2 == 0:
            s //= 2
        else:
            ans += tree[s]
            s = (s + 1) // 2
        if t % 2 == 1:
            t //= 2
        else:
            ans += tree[t]
            t = (t - 1) // 2
    return ans

def prime_factorize(n):
    ans = []
    for i in range(2, int(n ** (1 / 2)) + 1):
        while True:
            if n % i:
                break
            ans.append(i)
            n //= i
        if n == 1:
            break
    if not n == 1:
        ans.append(n)
    return ans

def euler_func(n):
    p = set(prime_factorize(n))
    x = n
    for i in p:
        x *= (i - 1)
        x //= i
    return x

def f(a, b):
    ans = 1
    for i in range(a + 1, b + 1):
        ans *= i
        ans %= mod
    return ans

n, mod = map(int, input().split())
x = euler_func(mod)
l = n + 5
fact = [1] * (l + 1)
for i in range(1, l + 1):
    fact[i] = i * fact[i - 1] % mod
if n <= 3:
    ans = 0
else:
    dp = [0] * 2000
    dp[0], dp[1], dp[2], dp[3] = 1, 2, 2, 1
    tree = make_tree(2005)
    for i in range(4):
        update(i, dp[i])
    m = 3
    ans = 0
    for i in range(n - 3):
        dp0 = list(dp)
        ans0 = 0
        for j in range(i + 3):
            for k in range(1, m + 2):
                ans0 += get_sum(j + k + 1, 2004) * dp0[k - 1] % mod
                ans0 %= mod
                dp[j + k] += dp0[k - 1]
                dp[j + k] %= mod
                update(j + k, dp[j + k])
        m += i + 3
        ans += f(i + 4, n) * ans0 % mod
        ans %= mod
print(ans)