#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;
typedef pair<int,P> P1;
typedef pair<P,P> P2;
#define pu push
#define pb push_back
#define mp make_pair
#define eps 1e-7
#define INF 1000000000
#define mod 1000000007
#define fi first
#define sc second
#define rep(i,x) for(int i=0;i<x;i++)
#define repn(i,x) for(int i=1;i<=x;i++)
#define SORT(x) sort(x.begin(),x.end())
#define ERASE(x) x.erase(unique(x.begin(),x.end()),x.end())
#define POSL(x,v) (lower_bound(x.begin(),x.end(),v)-x.begin())
#define POSU(x,v) (upper_bound(x.begin(),x.end(),v)-x.begin())
ll dp[105][8005];
int n,k;
ll modpow(ll x,ll n)
{
	ll res=1;
	while(n>0)
	{
		if(n&1) res=res*x%mod;
		x=x*x%mod;
		n>>=1;
	}
	return res;
}
int main(){
	cin>>n>>k;
	dp[n+1][0] = 1;
	for(int i=n+1;i>=2;i--){
		for(int j=0;j<=8000;j++){
			if(dp[i][j] == 0) continue;
			for(int nxt=0;nxt<=k;nxt++){
				int add = (j+nxt)/(i-1);
				if(nxt > i-1){
					dp[i-1][j] = (dp[i-1][j]+dp[i][j])%mod;
				}
				else{
					dp[i-1][j+add] = (dp[i-1][j+add]+dp[i][j])%mod;
				}
			}
		}
	}
	ll ret = 0;
	ret = 1LL*(k+1)*k/2LL*n%mod*modpow(k+1,n-1)%mod;
	for(int i=0;i<=8000;i++){
		ret -= dp[1][i] * 1LL * i%mod;
	}
	ret = (ret%mod+mod)%mod;
	cout<<ret<<endl;
}
### Prompt

In CPP, your task is to solve 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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;
typedef pair<int,P> P1;
typedef pair<P,P> P2;
#define pu push
#define pb push_back
#define mp make_pair
#define eps 1e-7
#define INF 1000000000
#define mod 1000000007
#define fi first
#define sc second
#define rep(i,x) for(int i=0;i<x;i++)
#define repn(i,x) for(int i=1;i<=x;i++)
#define SORT(x) sort(x.begin(),x.end())
#define ERASE(x) x.erase(unique(x.begin(),x.end()),x.end())
#define POSL(x,v) (lower_bound(x.begin(),x.end(),v)-x.begin())
#define POSU(x,v) (upper_bound(x.begin(),x.end(),v)-x.begin())
ll dp[105][8005];
int n,k;
ll modpow(ll x,ll n)
{
	ll res=1;
	while(n>0)
	{
		if(n&1) res=res*x%mod;
		x=x*x%mod;
		n>>=1;
	}
	return res;
}
int main(){
	cin>>n>>k;
	dp[n+1][0] = 1;
	for(int i=n+1;i>=2;i--){
		for(int j=0;j<=8000;j++){
			if(dp[i][j] == 0) continue;
			for(int nxt=0;nxt<=k;nxt++){
				int add = (j+nxt)/(i-1);
				if(nxt > i-1){
					dp[i-1][j] = (dp[i-1][j]+dp[i][j])%mod;
				}
				else{
					dp[i-1][j+add] = (dp[i-1][j+add]+dp[i][j])%mod;
				}
			}
		}
	}
	ll ret = 0;
	ret = 1LL*(k+1)*k/2LL*n%mod*modpow(k+1,n-1)%mod;
	for(int i=0;i<=8000;i++){
		ret -= dp[1][i] * 1LL * i%mod;
	}
	ret = (ret%mod+mod)%mod;
	cout<<ret<<endl;
}
```