#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const ll mod=1000000007;
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
// head

const int N=110;
void upd(int &a,int b) { a+=b; if (a>=mod) a-=mod; }
int n,m;
int dp[N][N*N];
ll ret;
int main() {
	scanf("%d%d",&n,&m);
	dp[n][0]=1;
	per(i,0,n) rep(j,0,n*m+1) if (dp[i+1][j]>0) {
		rep(k,0,m+1) {
			if (k<=i+1&&k+j>=i+1) upd(dp[i][j+(k+j)/(i+1)],dp[i+1][j]);
			else upd(dp[i][j],dp[i+1][j]);
		}
	}
	ret=m*(m+1)/2*n*powmod(m+1,n-1)%mod;
	rep(i,0,n*m+1) {
		ret=(ret-(ll)dp[0][i]*i)%mod;
	}
	if (ret<0) ret+=mod;
	printf("%lld\n",ret);
}

### 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;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const ll mod=1000000007;
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
// head

const int N=110;
void upd(int &a,int b) { a+=b; if (a>=mod) a-=mod; }
int n,m;
int dp[N][N*N];
ll ret;
int main() {
	scanf("%d%d",&n,&m);
	dp[n][0]=1;
	per(i,0,n) rep(j,0,n*m+1) if (dp[i+1][j]>0) {
		rep(k,0,m+1) {
			if (k<=i+1&&k+j>=i+1) upd(dp[i][j+(k+j)/(i+1)],dp[i+1][j]);
			else upd(dp[i][j],dp[i+1][j]);
		}
	}
	ret=m*(m+1)/2*n*powmod(m+1,n-1)%mod;
	rep(i,0,n*m+1) {
		ret=(ret-(ll)dp[0][i]*i)%mod;
	}
	if (ret<0) ret+=mod;
	printf("%lld\n",ret);
}

```