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 PyPy3 (2.4.0) (judge verdict: Accepted)

MOD = 10**9 + 7
n, oddness = [int(item) for item in input().split()]
# Oddness must be even
if oddness % 2 ==  1:
    print(0)
    exit()
oddness //= 2
max_oddness = n*n + n
max_pending = n + 1

dp = [[[0] * (max_oddness) for _ in range(max_pending)] for _ in range(n)]
oddness_itr = [item for item in range(n*n)]
for i in range(n):
    if i == 0:
        dp[0][0][0] = 1
        dp[0][1][1] = 1
        continue
    for j in range(i+1):
        for k in oddness_itr:
            # Connect in line
            # Oddness increase by num of pending item
            dp[i][j][k+j] += dp[i-1][j][k]
            # Choose 1 from pending
            dp[i][j][k+j] += dp[i-1][j][k] * j*2 
            dp[i][j][k+j] %= MOD
            # Choose 2 from pending
            if j > 0:
                dp[i][j-1][k+j-1] += dp[i-1][j][k] * j*j
                dp[i][j-1][k+j-1] %= MOD 
            # Do nothing 
            dp[i][j+1][k+j+1] += dp[i-1][j][k] 
            dp[i][j+1][k+j+1] %= MOD 

print(dp[n-1][0][oddness])