#include <iostream>
#include <fstream>
#include <cassert>
#include <typeinfo>
#include <vector>
#include <stack>
#include <cmath>
#include <set>
#include <map>
#include <string>
#include <algorithm>
#include <cstdio>
#include <queue>
#include <iomanip>
#include <cctype>
#include <random>
#include <complex>
#define syosu(x) fixed<<setprecision(x)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef pair<double,double> pdd;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<char> vc;
typedef vector<vc> vvc;
typedef vector<string> vs;
typedef vector<bool> vb;
typedef vector<vb> vvb;
typedef vector<P> vp;
typedef vector<vp> vvp;
typedef vector<pll> vpll;
typedef pair<int,P> pip;
typedef vector<pip> vip;
const int inf=1<<30;
const ll INF=1ll<<59;
const double pi=acos(-1);
const double eps=1e-7;
const ll mod=1e9+7;
const int dx[4]={0,1,0,-1},dy[4]={1,0,-1,0};

const int M=4000;

int n,k;

int main(){
	cin>>n>>k;
	vl dp,DP,dp1,DP1;
	dp=DP=dp1=DP1=vl(M);
	dp1[0]=1;
	int L=0;
	for(int i=n;i>0;i--){
		for(int j=0;j<M;j++) DP[j]=DP1[j]=0;
		for(int j=0;j<=k;j++) for(int l=0;l<=L;l++){
			if(j<=i){
				int I=l+(l+j)/i;
				(DP1[I]+=dp1[l])%=mod;
				(DP[I]+=dp[l]+(l+j)%i*dp1[l])%=mod;
			}
			else{
				(DP1[l]+=dp1[l])%=mod;
				(DP[l]+=dp[l]+(l+j)*dp1[l])%=mod;
			}
		}
		dp=DP;
		dp1=DP1;
		L+=(i+L)/i;
	}
	int res=0;
	for(int i=0;i<M;i++) (res+=dp[i])%=mod;
	cout<<res<<endl;
}
### Prompt

Please formulate a CPP solution to the following problem:

Consider the following game:

* The game is played using a row of N squares and many stones.
* First, a_i stones are put in Square i\ (1 \leq i \leq N).
* A player can perform the following operation as many time as desired: "Select an integer i such that Square i contains exactly i stones. Remove all the stones from Square i, and add one stone to each of the i-1 squares from Square 1 to Square i-1."
* The final score of the player is the total number of the stones remaining in the squares.



For a sequence a of length N, let f(a) be the minimum score that can be obtained when the game is played on a.

Find the sum of f(a) over all sequences a of length N where each element is between 0 and K (inclusive). Since it can be extremely large, find the answer modulo 1000000007 (= 10^9+7).

Constraints

* 1 \leq N \leq 100
* 1 \leq K \leq N

Input

Input is given from Standard Input in the following format:


N K


Output

Print the sum of f(a) modulo 1000000007 (= 10^9+7).

Examples

Input

2 2


Output

10


Input

20 17


Output

983853488

### Response

```cpp
#include <iostream>
#include <fstream>
#include <cassert>
#include <typeinfo>
#include <vector>
#include <stack>
#include <cmath>
#include <set>
#include <map>
#include <string>
#include <algorithm>
#include <cstdio>
#include <queue>
#include <iomanip>
#include <cctype>
#include <random>
#include <complex>
#define syosu(x) fixed<<setprecision(x)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef pair<double,double> pdd;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<char> vc;
typedef vector<vc> vvc;
typedef vector<string> vs;
typedef vector<bool> vb;
typedef vector<vb> vvb;
typedef vector<P> vp;
typedef vector<vp> vvp;
typedef vector<pll> vpll;
typedef pair<int,P> pip;
typedef vector<pip> vip;
const int inf=1<<30;
const ll INF=1ll<<59;
const double pi=acos(-1);
const double eps=1e-7;
const ll mod=1e9+7;
const int dx[4]={0,1,0,-1},dy[4]={1,0,-1,0};

const int M=4000;

int n,k;

int main(){
	cin>>n>>k;
	vl dp,DP,dp1,DP1;
	dp=DP=dp1=DP1=vl(M);
	dp1[0]=1;
	int L=0;
	for(int i=n;i>0;i--){
		for(int j=0;j<M;j++) DP[j]=DP1[j]=0;
		for(int j=0;j<=k;j++) for(int l=0;l<=L;l++){
			if(j<=i){
				int I=l+(l+j)/i;
				(DP1[I]+=dp1[l])%=mod;
				(DP[I]+=dp[l]+(l+j)%i*dp1[l])%=mod;
			}
			else{
				(DP1[l]+=dp1[l])%=mod;
				(DP[l]+=dp[l]+(l+j)*dp1[l])%=mod;
			}
		}
		dp=DP;
		dp1=DP1;
		L+=(i+L)/i;
	}
	int res=0;
	for(int i=0;i<M;i++) (res+=dp[i])%=mod;
	cout<<res<<endl;
}
```