#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <memory.h>
#include <math.h>
#include <assert.h>
#include <queue>
#include <map>
#include <set>
#include <string>
#include <algorithm>
#include <iostream>
#include <functional>
#include <unordered_map>
#include <unordered_set>
#include <list>
#include <bitset>

using namespace std;
typedef pair<int, int> pii;
typedef long long ll;
#define Fi first
#define Se second
#define pb(x) push_back(x)
#define szz(x) ((int)(x).size())
#define rep(i, n) for(int i=0;i<n;i++)
#define all(x) (x).begin(), (x).end()
typedef tuple<int, int, int> t3;
typedef pair<ll, ll> pll;
typedef long double ldouble;
typedef pair<double, double> pdd;

const int MOD = 1e9 + 7;
int pw(int x, int y) {
	int res = 1;
	while(y) {
		if(y & 1) res = (ll)res * x % MOD;
		x = (ll)x * x % MOD;
		y >>= 1;
	}
	return res;
}

int N, K;
int D[110][10010];

void solve() {
	scanf("%d%d", &N, &K);
	ll ans = K * (K + 1) / 2;
	ans = ans * pw(K+1, N-1) % MOD;
	ans = ans * N % MOD;
	D[N+1][0] = 1;
	int mx = 0;
	for(int i=N;i;i--) {
		for(int j=-i;j<=K-i;j++) {
			for(int k=0;k<=mx;k++) {
				int nk = k;
				if(j <= 0 && -j <= k) {
					nk = (k + (k + j) / i + 1);
					ans = (ans - pw(K+1, i-1) * (ll)D[i+1][k] % MOD * (nk - k)) % MOD;
					if(ans < 0) ans += MOD;
				}
				D[i][nk] = (D[i][nk] + D[i+1][k]) % MOD;
				if(D[i][nk]) mx = max(mx, nk);
			}
		}
	}
	printf("%lld\n", ans);
}

int main(){
	int Tc = 1; // scanf("%d\n", &Tc);
	for(int tc=1;tc<=Tc;tc++){
		solve();
	}
	return 0;
}

### 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 <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <memory.h>
#include <math.h>
#include <assert.h>
#include <queue>
#include <map>
#include <set>
#include <string>
#include <algorithm>
#include <iostream>
#include <functional>
#include <unordered_map>
#include <unordered_set>
#include <list>
#include <bitset>

using namespace std;
typedef pair<int, int> pii;
typedef long long ll;
#define Fi first
#define Se second
#define pb(x) push_back(x)
#define szz(x) ((int)(x).size())
#define rep(i, n) for(int i=0;i<n;i++)
#define all(x) (x).begin(), (x).end()
typedef tuple<int, int, int> t3;
typedef pair<ll, ll> pll;
typedef long double ldouble;
typedef pair<double, double> pdd;

const int MOD = 1e9 + 7;
int pw(int x, int y) {
	int res = 1;
	while(y) {
		if(y & 1) res = (ll)res * x % MOD;
		x = (ll)x * x % MOD;
		y >>= 1;
	}
	return res;
}

int N, K;
int D[110][10010];

void solve() {
	scanf("%d%d", &N, &K);
	ll ans = K * (K + 1) / 2;
	ans = ans * pw(K+1, N-1) % MOD;
	ans = ans * N % MOD;
	D[N+1][0] = 1;
	int mx = 0;
	for(int i=N;i;i--) {
		for(int j=-i;j<=K-i;j++) {
			for(int k=0;k<=mx;k++) {
				int nk = k;
				if(j <= 0 && -j <= k) {
					nk = (k + (k + j) / i + 1);
					ans = (ans - pw(K+1, i-1) * (ll)D[i+1][k] % MOD * (nk - k)) % MOD;
					if(ans < 0) ans += MOD;
				}
				D[i][nk] = (D[i][nk] + D[i+1][k]) % MOD;
				if(D[i][nk]) mx = max(mx, nk);
			}
		}
	}
	printf("%lld\n", ans);
}

int main(){
	int Tc = 1; // scanf("%d\n", &Tc);
	for(int tc=1;tc<=Tc;tc++){
		solve();
	}
	return 0;
}

```