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]. 
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int>pii;
typedef pair<long long,long long>pll;
ll dp[55][2000];//dp[i][j]表示长度为i的排列中逆序对为j的排列数
ll s[55][2000];//s[i][j]表示dp[i][0]到dp[i][j]的和
ll ans[55];//ans[i]代表长度为i的后半段满足条件的种数
ll n,p;
void init(){
    ll i,j;
    dp[0][0]=1;
    for(i=0;i<=n*(n-1)/2;i++) s[0][i]=1;
    for(i=1;i<=n;i++){//枚举第一个出现不同的位置
        for(j=0;j<=n*(n-1)/2;j++){
            if(j-i>=0) dp[i][j]=(s[i-1][j]-s[i-1][j-i]+p)%p;
            else dp[i][j]=s[i-1][j]%p;
            if(!j) s[i][j]=dp[i][j]%p;
            else s[i][j]=(s[i][j-1]+dp[i][j])%p;
        }
    }
}
void solve(){
    ll i,j,k,q,t;
    cin>>n>>p;
    init();
    for(i=1;i<=n;i++){//枚举后半段长度
        for(j=1;j<=i;j++){//枚举p
            for(k=j+1;k<=i;k++){//枚举q
                for(q=0;q<=i*(i-1)/2;q++){//统计逆序对数大于p-q的排列数
                    t=q-k+j;
                    if(t<=0) continue;
                    ans[i]=(ans[i]+dp[i-1][q]*s[i-1][t-1]%p)%p;
                }
            }
        }
    }
    for(i=2;i<=n;i++) ans[i]=(ans[i]+1ll*ans[i-1]*i)%p;//考虑前半段排列情况，将i-1归到i：C（i，1）
    printf("%lld",ans[n]);
}
int main(void){
    int T=1;
    //cin>>T;
    while(T--){
        solve();
    }
}