#include <iostream>
#include <cstring>
#include <cstdio>
typedef long long LL;

using namespace std;

inline int RD() {
    int x = 0, f = 1;
    char ch = getchar();

    while (ch < '0' || ch > '9') {
        if (ch == '-') f = -1;
        ch = getchar();
    }

    while (ch >= '0' && ch <= '9')
        x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}

const int N = 105;
const LL mod = 1e9 + 7;

#define MOD(x) (((x) + mod) % mod)
#define LF(i, __l, __r) for (int i = __l; i <= __r; i++)
#define RF(i, __r, __l) for (int i = __r; i >= __l; i--)

int n, k;
LL f[2][N][N * N]; // 第 1 维为处理到了第 i 个数，第 2 维已经处理好了 j 对数，第 3 维为现在数的和。
// 表示方案数

int main() {
    n = RD(), k = RD();
    f[0][0][0] = 1;

    LF(i, 1, n) {
        memset(f[i & 1], 0, sizeof(f[i & 1]));
        LF(j, 0, i - 1) {
            LF(l, 0, k) {
                LL x = j, y = (i - 1) - j, z = n - x - y * 2;
                LL &a = f[i & 1][j + 1][l + i],
                    &b = f[i & 1][j + 2][l + 2 * i], 
                    &c = f[i & 1][j][l],
                    d = f[(i - 1) & 1][j][l];
                a = MOD(MOD(d * z) + a);
                a = MOD(MOD(d * MOD(y * MOD(2 * z))) + a);
                b = MOD(MOD(d * y) * y + b);
                c = MOD(MOD(d * z) * (z - 1) + c);
            }
        }
    }

    printf("%lld", f[n & 1][n][k]);
    return 0;
}