#include <bits/stdc++.h>
using namespace std;
typedef long long lint;
typedef long double llf;
typedef pair<lint, lint> pi;
const int MAXN = 200005;
const int mod = 1e9 + 7;

int n, k;
bool vis[105][10005];
pi dp[105][10005];

int mx = 0;
pi f(int cur, int exp){
	mx = max(mx, exp);
	if(cur == 0) return pi(0, 1);
	if(vis[cur][exp]) return dp[cur][exp];
	pi ans(0, 0);
	for(int i=0; i<=k; i++){
		if(i > cur){
			auto x = f(cur - 1, exp);
			ans.first += x.first + x.second * (i + exp) % mod;
			ans.second += x.second;
		}
		else{
			auto x = f(cur - 1, exp + (exp + i) / cur);
			ans.first += x.first + x.second * ((exp + i) % cur);
			ans.second += x.second;
		}
	}
	ans.first %= mod;
	ans.second %= mod;
	vis[cur][exp] = 1;
	return dp[cur][exp] = ans;
}

int main(){
	cin >> n >> k;
	cout << f(n, 0).first<< endl;
}

### Prompt

Your task is to create a Cpp solution to 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 lint;
typedef long double llf;
typedef pair<lint, lint> pi;
const int MAXN = 200005;
const int mod = 1e9 + 7;

int n, k;
bool vis[105][10005];
pi dp[105][10005];

int mx = 0;
pi f(int cur, int exp){
	mx = max(mx, exp);
	if(cur == 0) return pi(0, 1);
	if(vis[cur][exp]) return dp[cur][exp];
	pi ans(0, 0);
	for(int i=0; i<=k; i++){
		if(i > cur){
			auto x = f(cur - 1, exp);
			ans.first += x.first + x.second * (i + exp) % mod;
			ans.second += x.second;
		}
		else{
			auto x = f(cur - 1, exp + (exp + i) / cur);
			ans.first += x.first + x.second * ((exp + i) % cur);
			ans.second += x.second;
		}
	}
	ans.first %= mod;
	ans.second %= mod;
	vis[cur][exp] = 1;
	return dp[cur][exp] = ans;
}

int main(){
	cin >> n >> k;
	cout << f(n, 0).first<< endl;
}

```