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;
#define endl "\n"
#define mst(a,b) memset((a),(b),sizeof(a))
#define rush() int TT;scanf("%d",&TT);for(int TTT=1;TTT<=TT;TTT++)
#define fastio() ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define forrr(i,x,y) for(int i=x;i<=y;++i)
#define rofff(i,x,y) for(int i=x;i>=y;--i)
#define forr(i,x,y) for(int i=x;i<y;++i)
#define roff(i,x,y) for(int i=x;i>y;--i)
#define lowbit(x) (x&-x)
#define fin freopen("in.txt","r",stdin)
#define fout freopen("out.txt","w",stdout)
#define fi first
#define se second
#define mp make_pair
#define pb emplace_back
#define sf scanf
#define pf printf
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ldb;
typedef pair<int, int> pii;
typedef pair<ll, int> pli;
typedef pair<ll, ll> pll;
typedef pair<string,int> psi;
const ldb pi = acos((ldb)-1);
int in(int &a,int &b,int &c,int &d) {
    return scanf("%d%d%d%d",&a,&b,&c,&d);
}
int in(int &a,int &b,int &c) {
    return scanf("%d%d%d",&a,&b,&c);
}
int in(int &a,int &b) {
    return scanf("%d%d",&a,&b);
}
int in(ll &a,ll &b) {
    return scanf("%lld%lld",&a,&b);
}
int in(ll &a,ll &b,ll &c,ll &d) {
    return scanf("%lld%lld%lld%lld",&a,&b,&c,&d);
}
int in(ll &a,ll &b,ll &c) {
    return scanf("%lld%lld%lld",&a,&b,&c);
}
int in(ll &a) {
    return scanf("%lld",&a);
}
int in(int &a) {
    return scanf("%d",&a);
}
int in(char *s) {
    return scanf("%s",s);
}
int in(char &c) {
    return scanf("%c",&c);
}
void out(int a,bool ln) {
    printf("%d%c",a," \n"[ln]);
}
void out(ll a,bool ln) {
    printf("%lld%c",a," \n"[ln]);
}
void out(db a,int digit,bool ln) {
    printf("%.*f%c",digit,a," \n"[ln]);
}
ll gcd(ll a, ll b) {
    return b?gcd(b,a%b):a;
}
ll lcm(ll a, ll b) {
    return a/gcd(a,b)*b;
}
const int inf = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 3e5+5;
const int maxm = 1e6+5;
const ll mod = 1e9+7;
ll dp[55][2500],pre[55][2500],ans[55],n,m;

int main() {
    in(n,m);
    dp[0][0]=1;
    forrr(i,0,n*(n-1)/2) {
        pre[0][i]=1;
    }
    forrr(i,1,n) {
        dp[i][0]=pre[i][0]=1;
        forrr(j,1,n*(n-1)/2) {
            if(j-i>=0)
                dp[i][j]=(pre[i-1][j]-pre[i-1][j-i]+m)%m;
            else
                dp[i][j]=pre[i-1][j];
            pre[i][j]=(pre[i][j-1]+dp[i][j])%m;
        }
    }
    forrr(i,1,n) {
        forrr(j,1,i) {
            forr(k,1,j) {
                forrr(l,0,(i-1)*(i-2)/2) {
                    if(l+k-j-1>=0)
                        ans[i]=(ans[i]+dp[i-1][l]*pre[i-1][l+k-j-1]%m)%m;
                }
            }
        }
        ans[i] = (1LL*i*ans[i-1]%m+ans[i])%m;
    }
    out(ans[n],1);
}
