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++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
#define SZ(x) ((int)x.size())
#define UNI(x) sort(ALL(x)),x.resize(unique(ALL(x))-x.begin());
#define GETPOS(c,x) (lower_bound(ALL(c),x)-c.begin())
#define GETpos(c,x) (lower_bound(ALL(c),x)-c.begin())////
#define MST(x,val) memset((x),val,sizeof((x)));
#define ls rt<<1
#define rs rt<<1|1
#define lson rt<<1,l,M
#define rson rt<<1|1,M+1,r
#define PI acos(-1)
#define MM int M=(l+r)>>1;
#define fu(i,r,t) for(int i=r;i<=t;i++)
#define fd(i,r,t) for(int i=r;i>=t;i--)
#define fh(i,be,e) for(int i=head[be];~i;i=e[i].next)
#define fa(i,V) for(auto i:V)
#define far(i,V) for(auto &i:V)
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define int long long
#define endl '\n'
#define lcm(a,b) (a*b)/__gcd(a,b)
#define cp(i,ans) printf("%.if",ans);
#define cpp(i,ans) cout<<setprecision(i)<<fixed<<ans<<endl;
#define ppb pop_back
#define ppf pop_front
#define pb push_back
#define pf push_front
#define pq priority_queue
#define lowbit(x) (x)&(-x)
#define all(V) V.begin(),V.end()
#define ms multiset
#define mod(x) ((x+mo_num)%mo_num)
#define vc vector
#define vct vector<int>
#define out(i) cout<<(i)<<endl;
#define fi first
#define se second
#define fun(i) fu(i,1,n)
#define fut(i) fu(i,1,t)
#define fum(i) fu(i,1,m)
#define ld long double
#define umap unordered_map
#define P pair<int,int>
#define SET set<int>
#define mp map<int,int>
#define mk make_tuple
#define low lower_bound
#define upp upper_bound
#define yn(key) out(key?"YES":"NO")
#define in(i) i+1,i+1+n
#define im(i) i+1,i+1+m
#define bffs(i) __builtin_ffs(i)
#define bcount(i) __builtin_popcount(i)
#define bone(i) ((1<<i)-1)
#define db double
#define got(container,num) get<num-1>(container)
using namespace std;
//Remember cancel"#define endl '\n'" in interactive questions or use "<<flush"
// use C:printf("%.16f", x);  ->  printf("%.10f", x); can accelerate the program
typedef unsigned long long ull; template<class T> void _R(T &x) { cin >> x; }void _R(int32_t &x) { scanf("%d", &x); }void _R(long long &x) { scanf("%lld", &x); }void _R(ull &x) { scanf("%llu", &x); }void _R(double &x) { scanf("%lf", &x); }void _R(char &x) { scanf(" %c", &x); }void _R(char *x) { scanf("%s", x); }void R() {}template<class T, class... U> void R(T &head, U &... tail) { _R(head); R(tail...); }template<class T> void _W(const T &x) { cout << x; }void _W(const int32_t &x) { printf("%d", x); }void _W(const long long &x) { printf("%lld", x); }void _W(const double &x) { printf("%.16f", x); }void _W(const char &x) { putchar(x); }void _W(const char *x) { printf("%s", x); }template<class T,class U> void _W(const pair<T,U> &x) {_W(x.F); putchar(' '); _W(x.S);}template<class T> void _W(const vector<T> &x) { for (auto i = x.begin(); i != x.end(); _W(*i++)) if (i != x.cbegin()) putchar(' '); }void W() {}template<class T, class... U> void W(const T &head, const U &... tail) { _W(head); putchar(sizeof...(tail) ? ' ' : '\n'); W(tail...); }
const int INF=LLONG_MAX/2,SINF=0x3f3f3f3f,Lim=1<<20,MINF=LLONG_MAX;
const int dx[]={0,0,-1,1},dy[]={-1,1,0,0};//up down left right
const int maxn=1e6+1e5;
const int mo_num=1e9+7;
//const int mo_num=998244353;
int n,m,t,a[maxn],b[maxn],ans;

const int N=101;
int f[N][N][N*N];
void add(int &q,int w)
{
    q=mod(q+w);
    return;
}
void solve()
{
    cin>>n>>m;
    f[1][0][0]=1;
    f[1][1][1]=1;
    fun(i)
    {
        fu(j,0,i)
        {
            fu(k,0,m)
            {
                add(f[i+1][j][k+j],mod((2*j+1)*f[i][j][k]));
                if(j)add(f[i+1][j-1][k+j-1],mod(mod(j*j)*f[i][j][k]));
                add(f[i+1][j+1][k+j+1],f[i][j][k]);
            }
        }
    }
    out(m%2?0:f[n][0][m/2])
    return ;
}
main()
{
    IOS
    int T=1;
    //cin>>T;
    while(T--)solve();
    return 0;
}