#include <cstdio>
#include <iostream>
#include <algorithm>
#include <string>
#include <cstring>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <cmath>
#include <iomanip>
#include <cassert>
#include <bitset>
using namespace std;

typedef pair<int, int> P;
#define rep(i, n) for (int i=0; i<(n); i++)
#define all(c) (c).begin(), (c).end()
#define uniq(c) c.erase(unique(all(c)), (c).end())
#define index(xs, x) (int)(lower_bound(all(xs), x) - xs.begin())
#define _1 first
#define _2 second
#define pb push_back
#define INF 1145141919
#define MOD 1000000007
inline void add(int &x, int v) { x += v; if (x >= MOD) x -= MOD; }
inline long long mul(int x, int y) { return (1LL*x*y) % MOD; }
const int B = 10001;

int N, K;
int pK[101];
int dp[2][B];

signed main() {
  ios::sync_with_stdio(false); cin.tie(0);
  cin >> N >> K;
  pK[0] = 1;
  for (int i=1; i<=N; i++) pK[i] = (1LL*(K+1)*pK[i-1]) % MOD;
  int ans = 0;
  dp[0][0] = 1;
  for (int i=N; i>=1; i--) {
    rep(j, B) dp[1][j] = 0;
    for (int s=B-1; s>=0; s--) {
      if (dp[0][s] == 0) continue;
      for (int t=0; t<=K; t++) {
        if (t > i) {
          add(dp[1][s], dp[0][s]);
          add(ans, mul(pK[i-1], mul(dp[0][s], t+s)));
        }
        else {
          add(dp[1][s+(t+s)/i], dp[0][s]);
          add(ans, mul(pK[i-1], mul(dp[0][s], (t+s)%i)));
        }
      }
    }
    swap(dp[0], dp[1]);
  }
  cout << ans << "\n";
  return 0;
}

### Prompt

Create a solution in CPP for 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 <cstdio>
#include <iostream>
#include <algorithm>
#include <string>
#include <cstring>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <cmath>
#include <iomanip>
#include <cassert>
#include <bitset>
using namespace std;

typedef pair<int, int> P;
#define rep(i, n) for (int i=0; i<(n); i++)
#define all(c) (c).begin(), (c).end()
#define uniq(c) c.erase(unique(all(c)), (c).end())
#define index(xs, x) (int)(lower_bound(all(xs), x) - xs.begin())
#define _1 first
#define _2 second
#define pb push_back
#define INF 1145141919
#define MOD 1000000007
inline void add(int &x, int v) { x += v; if (x >= MOD) x -= MOD; }
inline long long mul(int x, int y) { return (1LL*x*y) % MOD; }
const int B = 10001;

int N, K;
int pK[101];
int dp[2][B];

signed main() {
  ios::sync_with_stdio(false); cin.tie(0);
  cin >> N >> K;
  pK[0] = 1;
  for (int i=1; i<=N; i++) pK[i] = (1LL*(K+1)*pK[i-1]) % MOD;
  int ans = 0;
  dp[0][0] = 1;
  for (int i=N; i>=1; i--) {
    rep(j, B) dp[1][j] = 0;
    for (int s=B-1; s>=0; s--) {
      if (dp[0][s] == 0) continue;
      for (int t=0; t<=K; t++) {
        if (t > i) {
          add(dp[1][s], dp[0][s]);
          add(ans, mul(pK[i-1], mul(dp[0][s], t+s)));
        }
        else {
          add(dp[1][s+(t+s)/i], dp[0][s]);
          add(ans, mul(pK[i-1], mul(dp[0][s], (t+s)%i)));
        }
      }
    }
    swap(dp[0], dp[1]);
  }
  cout << ans << "\n";
  return 0;
}

```