#include <iostream>
#include <algorithm>
#include <vector>
#define FOR(i,a,b) for(ut i=(a);i<(int)(b);i++)
#define REP(i,b) FOR(i,0,b)
#define ALL(c) c.begin(),c.end()
#define PB push_back
#define cat //cout << __LINE__ << endl;
using namespace std;
typedef long long LL;
typedef double ld;
typedef LL ut;
typedef vector<ut> VI;
typedef pair<ut,ut> pr;
typedef pair<ut,pr> ppr;
typedef vector<pr> Vpr;
typedef vector<ppr> Vppr;
using namespace std;
const int SIZE=105;
const LL p=7+1e9;
const LL INF=1LL<<60;
LL DP[SIZE][SIZE*SIZE*2];
LL N,K;
LL power(LL n,LL r){
	if(r==0) return 1;
	if(r==1) return n%p;
	LL ans=power(n,r/2);
	return ans*ans%p*power(n,r%2)%p;
}
LL als(LL n,LL k){
	LL pls=k*(k+1)/2%p*power(k+1,n-1)%p*n%p;
	return pls;
}
LL solve(){
	LL ans=0;
	cin >>N >>K;
	DP[N][0]=1;
	REP(i,N){
		LL now=N-i;
		LL times=0;
		REP(j,SIZE*SIZE/now){
			REP(k,K+1){
				times=0;
				if(now>=k){
					times=(j+k)/now;
				}
				DP[now-1][j+times]+=DP[now][j]%p;
					
			}
		}
	}
	REP(i,SIZE*SIZE){
		ans+=(LL)i*DP[0][i];
		ans%=p;
	}
	return (als(N,K)+p-ans)%p;
}
int main(){
	cout << solve() << endl;


	return 0;
}
### Prompt

Develop a solution in cpp to the problem described below:

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 <algorithm>
#include <vector>
#define FOR(i,a,b) for(ut i=(a);i<(int)(b);i++)
#define REP(i,b) FOR(i,0,b)
#define ALL(c) c.begin(),c.end()
#define PB push_back
#define cat //cout << __LINE__ << endl;
using namespace std;
typedef long long LL;
typedef double ld;
typedef LL ut;
typedef vector<ut> VI;
typedef pair<ut,ut> pr;
typedef pair<ut,pr> ppr;
typedef vector<pr> Vpr;
typedef vector<ppr> Vppr;
using namespace std;
const int SIZE=105;
const LL p=7+1e9;
const LL INF=1LL<<60;
LL DP[SIZE][SIZE*SIZE*2];
LL N,K;
LL power(LL n,LL r){
	if(r==0) return 1;
	if(r==1) return n%p;
	LL ans=power(n,r/2);
	return ans*ans%p*power(n,r%2)%p;
}
LL als(LL n,LL k){
	LL pls=k*(k+1)/2%p*power(k+1,n-1)%p*n%p;
	return pls;
}
LL solve(){
	LL ans=0;
	cin >>N >>K;
	DP[N][0]=1;
	REP(i,N){
		LL now=N-i;
		LL times=0;
		REP(j,SIZE*SIZE/now){
			REP(k,K+1){
				times=0;
				if(now>=k){
					times=(j+k)/now;
				}
				DP[now-1][j+times]+=DP[now][j]%p;
					
			}
		}
	}
	REP(i,SIZE*SIZE){
		ans+=(LL)i*DP[0][i];
		ans%=p;
	}
	return (als(N,K)+p-ans)%p;
}
int main(){
	cout << solve() << endl;


	return 0;
}
```