#include <bits/stdc++.h>
using namespace std;

typedef pair<int, int> pii;
typedef long long ll;
typedef vector<int> vi;

#define pb push_back
#define eb emplace_back
#define mp make_pair
#define fi first
#define se second
#define rep(i,n) rep2(i,0,n)
#define rep2(i,m,n) for(int i=m;i<(n);i++)
#define ALL(c) (c).begin(),(c).end()

const ll MOD = 1000000007;

int N, K;
ll dp[110][10010];

int main() {
	cin >> N >> K;
	dp[N][0] = 1;
	for (int i = N-1; i>=0; --i) {
		rep(j, 10010) if (dp[i+1][j] != 0) {
			for (int k = 0; k <= K; ++k) {
				int t = (i+1)-k;

				if (t < 0 || t > j) {
					dp[i][j] = (dp[i][j] + dp[i+1][j]) % MOD;
				} else {
					int num = (j-t)/(i+1) + 1;
					dp[i][j+num] = (dp[i][j+num] + dp[i+1][j]) % MOD;
				}
			}
		}
	}

	ll ret = N * K * (K+1) / 2;
	rep(i, N-1) ret = ret * (K+1) % MOD;

	rep(i, 10010) if (dp[0][i]>0) {
		ret = (ret - dp[0][i] * i % MOD);
		if (ret < 0) ret += MOD;
	}

	cout << ret << 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 <bits/stdc++.h>
using namespace std;

typedef pair<int, int> pii;
typedef long long ll;
typedef vector<int> vi;

#define pb push_back
#define eb emplace_back
#define mp make_pair
#define fi first
#define se second
#define rep(i,n) rep2(i,0,n)
#define rep2(i,m,n) for(int i=m;i<(n);i++)
#define ALL(c) (c).begin(),(c).end()

const ll MOD = 1000000007;

int N, K;
ll dp[110][10010];

int main() {
	cin >> N >> K;
	dp[N][0] = 1;
	for (int i = N-1; i>=0; --i) {
		rep(j, 10010) if (dp[i+1][j] != 0) {
			for (int k = 0; k <= K; ++k) {
				int t = (i+1)-k;

				if (t < 0 || t > j) {
					dp[i][j] = (dp[i][j] + dp[i+1][j]) % MOD;
				} else {
					int num = (j-t)/(i+1) + 1;
					dp[i][j+num] = (dp[i][j+num] + dp[i+1][j]) % MOD;
				}
			}
		}
	}

	ll ret = N * K * (K+1) / 2;
	rep(i, N-1) ret = ret * (K+1) % MOD;

	rep(i, 10010) if (dp[0][i]>0) {
		ret = (ret - dp[0][i] * i % MOD);
		if (ret < 0) ret += MOD;
	}

	cout << ret << endl;

	return 0;
}
```