Problem:
Score : 600 points

Problem Statement

Let us define the oddness of a permutation p = {p_1,\ p_2,\ ...,\ p_n} of {1,\ 2,\ ...,\ n} as \sum_{i = 1}^n |i - p_i|.

Find the number of permutations of {1,\ 2,\ ...,\ n} of oddness k, modulo 10^9+7.

Constraints

All values in input are integers.

1 \leq n \leq 50

0 \leq k \leq n^2

Input

Input is given from Standard Input in the following format:

n k

Output

Print the number of permutations of {1,\ 2,\ ...,\ n} of oddness k, modulo 10^9+7.

Sample Input 1

3 2

Sample Output 1

2

There are six permutations of {1,\ 2,\ 3}. Among them, two have oddness of 2: {2,\ 1,\ 3} and {1,\ 3,\ 2}.

Sample Input 2

39 14

Sample Output 2

74764168

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
typedef long long ll;

// auto mod int
// https://youtu.be/L8grWxBlIZ4?t=9858
// https://youtu.be/ERZuLAxZffQ?t=4807 : optimize
// https://youtu.be/8uowVvQ_-Mo?t=1329 : division
const int mod = 1000000007;
struct mint {
  ll x; // typedef long long ll;
  mint(ll x=0):x(x%mod){}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) {
    (x *= a.x) %= mod;
    return *this;
  }
  mint operator+(const mint a) const {
    mint res(*this);
    return res+=a;
  }
  mint operator-(const mint a) const {
    mint res(*this);
    return res-=a;
  }
  mint operator*(const mint a) const {
    mint res(*this);
    return res*=a;
  }
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }

  // for prime mod
  mint inv() const {
    return pow(mod-2);
  }
  mint& operator/=(const mint a) {
    return (*this) *= a.inv();
  }
  mint operator/(const mint a) const {
    mint res(*this);
    return res/=a;
  }
};

mint dp[55][55][2605];

int main() {
  int n, score;
  cin >> n >> score;
  dp[0][0][0] = 1;
  rep(i,n) {
    rep(j,i+1)rep(k,score+1) {
      {
        int nj = j;
        dp[i+1][nj][k+nj*2] += dp[i][j][k];
      }
      {
        int nj = j+1;
        dp[i+1][nj][k+nj*2] += dp[i][j][k];
      }
      {
        int nj = j;
        mint x = j*2;
        dp[i+1][nj][k+nj*2] += dp[i][j][k]*x;
      }
      if (j >= 1) {
        int nj = j-1;
        mint x = j*j;
        dp[i+1][nj][k+nj*2] += dp[i][j][k]*x;
      }
    }
  }
  mint ans = dp[n][0][score];
  cout << ans.x << endl;
  return 0;
}