#include <bits/stdc++.h>
#ifdef DEMETRIO
#define deb(...) fprintf(stderr,__VA_ARGS__)
#define deb1(x) cerr << #x << " = " << x << endl
#else
#define deb(...) 0
#define deb1(x) 0
#endif
#define pb push_back
#define mp make_pair
#define fst first
#define snd second
#define fore(i,a,b) for(int i=a,ThxDem=b;i<ThxDem;++i)
#define SZ(x) ((int)x.size())
#define MOD 1000000007
using namespace std;
typedef long long ll;

int add(int x, int y){return (x+y)%MOD;}
int mul(int x, int y){return (1LL*x*y)%MOD;}

int dp[128][16384];
int pk1[128];
int n,k;

int f(int n, int o){
	if(dp[n][o]>=0)return dp[n][o];
	int r=0;
	fore(j,0,k+1){
		if(j>n){r=add(r,add(mul(pk1[n-1],j+o),f(n-1,o)));}
		else {
			r=add(r,add(mul((j+o)%n,pk1[n-1]),f(n-1,o+(j+o)/n)));
		}
	}
	//printf("%d %d %d\n",n,o,r);
	return dp[n][o]=r;
}

int main(){
	scanf("%d%d",&n,&k);
	pk1[0]=1;
	fore(i,1,128)pk1[i]=mul(pk1[i-1],k+1);
	memset(dp,-1,sizeof(dp));
	fill(dp[0],dp[0]+16384,0);
	printf("%d\n",f(n,0));
	return 0;
}
### Prompt

Please provide a CPP coded solution 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 <bits/stdc++.h>
#ifdef DEMETRIO
#define deb(...) fprintf(stderr,__VA_ARGS__)
#define deb1(x) cerr << #x << " = " << x << endl
#else
#define deb(...) 0
#define deb1(x) 0
#endif
#define pb push_back
#define mp make_pair
#define fst first
#define snd second
#define fore(i,a,b) for(int i=a,ThxDem=b;i<ThxDem;++i)
#define SZ(x) ((int)x.size())
#define MOD 1000000007
using namespace std;
typedef long long ll;

int add(int x, int y){return (x+y)%MOD;}
int mul(int x, int y){return (1LL*x*y)%MOD;}

int dp[128][16384];
int pk1[128];
int n,k;

int f(int n, int o){
	if(dp[n][o]>=0)return dp[n][o];
	int r=0;
	fore(j,0,k+1){
		if(j>n){r=add(r,add(mul(pk1[n-1],j+o),f(n-1,o)));}
		else {
			r=add(r,add(mul((j+o)%n,pk1[n-1]),f(n-1,o+(j+o)/n)));
		}
	}
	//printf("%d %d %d\n",n,o,r);
	return dp[n][o]=r;
}

int main(){
	scanf("%d%d",&n,&k);
	pk1[0]=1;
	fore(i,1,128)pk1[i]=mul(pk1[i-1],k+1);
	memset(dp,-1,sizeof(dp));
	fill(dp[0],dp[0]+16384,0);
	printf("%d\n",f(n,0));
	return 0;
}
```