#include<stdio.h>
#include<iostream>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<utility>
#include<stack>
#include<algorithm>
#include<string>
#include<string.h>
#define fi first
#define se second
#define pb push_back
using namespace std;
typedef long long ll;
typedef pair<ll,ll> ii;
const int N = 100;
const int LIM = 4000;
const int MOD = 1e9 + 7;
ll dp[N + 5][LIM + 5];
ll png[N + 5];

int main(){
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    ll n, k; cin >> n >> k;
    png[0] = 1;
    for(int i = 1; i <= n; ++i){
        png[i] = 1LL * png[i - 1] * (k + 1) % MOD;
    }
    dp[n + 1][0] = 1;
    // cout << "TES" << endl;
    ll ans = 0;
    ll cnt = 0;
    for(int i = n; i >= 1; --i){
        for(int j = 0; j <= k; ++j){
            for(int l = 0; l <= cnt; ++l){
                ll val = j + l;

                if(j <= i){
                    dp[i][l + val / i] += dp[i + 1][l];
                    dp[i][l + val / i] %= MOD;
                    val %= i;
                }else{
                    dp[i][l] += dp[i + 1][l];
                    dp[i][l] %= MOD;
                }
                ll add = val * dp[i + 1][l] % MOD * png[i - 1] % MOD;
                // cout << "(" << i << ", " << j << ", " << l << ") -> " << add << " " << dp[i + 1][l] << endl;
                ans += add;
                ans %= MOD;
            }
        }
        cnt += (i + cnt) / i;
        // cout << i << " :: ";
        // for(int j = 0; j <= 10; ++j){
        //     cout << dp[i][j] << " ";
        // }cout << endl;
    }

    cout << ans << 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<stdio.h>
#include<iostream>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<utility>
#include<stack>
#include<algorithm>
#include<string>
#include<string.h>
#define fi first
#define se second
#define pb push_back
using namespace std;
typedef long long ll;
typedef pair<ll,ll> ii;
const int N = 100;
const int LIM = 4000;
const int MOD = 1e9 + 7;
ll dp[N + 5][LIM + 5];
ll png[N + 5];

int main(){
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    ll n, k; cin >> n >> k;
    png[0] = 1;
    for(int i = 1; i <= n; ++i){
        png[i] = 1LL * png[i - 1] * (k + 1) % MOD;
    }
    dp[n + 1][0] = 1;
    // cout << "TES" << endl;
    ll ans = 0;
    ll cnt = 0;
    for(int i = n; i >= 1; --i){
        for(int j = 0; j <= k; ++j){
            for(int l = 0; l <= cnt; ++l){
                ll val = j + l;

                if(j <= i){
                    dp[i][l + val / i] += dp[i + 1][l];
                    dp[i][l + val / i] %= MOD;
                    val %= i;
                }else{
                    dp[i][l] += dp[i + 1][l];
                    dp[i][l] %= MOD;
                }
                ll add = val * dp[i + 1][l] % MOD * png[i - 1] % MOD;
                // cout << "(" << i << ", " << j << ", " << l << ") -> " << add << " " << dp[i + 1][l] << endl;
                ans += add;
                ans %= MOD;
            }
        }
        cnt += (i + cnt) / i;
        // cout << i << " :: ";
        // for(int j = 0; j <= 10; ++j){
        //     cout << dp[i][j] << " ";
        // }cout << endl;
    }

    cout << ans << endl;
}
```