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

using VI = vector<int>;
using VVI = vector<VI>;
using PII = pair<int, int>;
using LL = long long;
using VL = vector<LL>;
using VVL = vector<VL>;
using PLL = pair<LL, LL>;
using VS = vector<string>;

#define ALL(a)  begin((a)),end((a))
#define RALL(a) (a).rbegin(), (a).rend()
#define PB push_back
#define EB emplace_back
#define MP make_pair
#define SZ(a) int((a).size())
#define SORT(c) sort(ALL((c)))
#define RSORT(c) sort(RALL((c)))
#define UNIQ(c) (c).erase(unique(ALL((c))), end((c)))

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)

#define FF first
#define SS second
template<class S, class T>
istream& operator>>(istream& is, pair<S,T>& p){
  return is >> p.FF >> p.SS;
}
template<class S, class T>
ostream& operator<<(ostream& os, const pair<S,T>& p){
  return os << p.FF << " " << p.SS;
}
template<class T>
void maxi(T& x, T y){
  if(x < y) x = y;
}
template<class T>
void mini(T& x, T y){
  if(x > y) x = y;
}


const double EPS = 1e-10;
const double PI  = acos(-1.0);
const LL MOD = 1e9+7;

const int MX = 3500;
LL dp[110][MX];

LL powm(LL x, LL y){
  if(y == 0)
	return 1;
  return powm(x*x%MOD, y/2) * (y%2?x:1) % MOD;
}
LL inv(LL x){
  return powm(x, MOD-2);
}

int main(){
  cin.tie(0);
  ios_base::sync_with_stdio(false);

  LL N, K;
  cin >> N >> K;

  dp[N+1][0] = 1;
  for(int i=N;i>=1;--i){
	for(int k=0;k<=K;++k){
	  if(i < k){
		for(int t=0;t<MX;++t){
		  (dp[i][t] += dp[i+1][t]) %= MOD;
		}
	  }
	  else{
		for(int t=0;t+(k+t)/i<MX;++t){
		  LL t_ = (k + t) / i;
		  (dp[i][t+t_] += dp[i+1][t]) %= MOD;
		}
	  }
	}
  }
  LL ans = K * powm(K+1, N) % MOD * inv(2) % MOD * N % MOD;
  REP(t,MX) (ans += MOD - dp[1][t] * t % MOD) %= MOD;
  
  cout << ans << endl;

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

using VI = vector<int>;
using VVI = vector<VI>;
using PII = pair<int, int>;
using LL = long long;
using VL = vector<LL>;
using VVL = vector<VL>;
using PLL = pair<LL, LL>;
using VS = vector<string>;

#define ALL(a)  begin((a)),end((a))
#define RALL(a) (a).rbegin(), (a).rend()
#define PB push_back
#define EB emplace_back
#define MP make_pair
#define SZ(a) int((a).size())
#define SORT(c) sort(ALL((c)))
#define RSORT(c) sort(RALL((c)))
#define UNIQ(c) (c).erase(unique(ALL((c))), end((c)))

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)

#define FF first
#define SS second
template<class S, class T>
istream& operator>>(istream& is, pair<S,T>& p){
  return is >> p.FF >> p.SS;
}
template<class S, class T>
ostream& operator<<(ostream& os, const pair<S,T>& p){
  return os << p.FF << " " << p.SS;
}
template<class T>
void maxi(T& x, T y){
  if(x < y) x = y;
}
template<class T>
void mini(T& x, T y){
  if(x > y) x = y;
}


const double EPS = 1e-10;
const double PI  = acos(-1.0);
const LL MOD = 1e9+7;

const int MX = 3500;
LL dp[110][MX];

LL powm(LL x, LL y){
  if(y == 0)
	return 1;
  return powm(x*x%MOD, y/2) * (y%2?x:1) % MOD;
}
LL inv(LL x){
  return powm(x, MOD-2);
}

int main(){
  cin.tie(0);
  ios_base::sync_with_stdio(false);

  LL N, K;
  cin >> N >> K;

  dp[N+1][0] = 1;
  for(int i=N;i>=1;--i){
	for(int k=0;k<=K;++k){
	  if(i < k){
		for(int t=0;t<MX;++t){
		  (dp[i][t] += dp[i+1][t]) %= MOD;
		}
	  }
	  else{
		for(int t=0;t+(k+t)/i<MX;++t){
		  LL t_ = (k + t) / i;
		  (dp[i][t+t_] += dp[i+1][t]) %= MOD;
		}
	  }
	}
  }
  LL ans = K * powm(K+1, N) % MOD * inv(2) % MOD * N % MOD;
  REP(t,MX) (ans += MOD - dp[1][t] * t % MOD) %= MOD;
  
  cout << ans << endl;

  return 0;
}

```