#include <bits/stdc++.h>
#include <unistd.h>

using namespace std;

#define DEBUG(x) cerr<<#x<<": "<<x<<endl;
#define DEBUG_VEC(v) cerr<<#v<<":";for(int i=0;i<v.size();i++) cerr<<" "<<v[i]; cerr<<endl
#define DEBUG_MAT(v) cerr<<#v<<endl;for(int i=0;i<v.size();i++){for(int j=0;j<v[i].size();j++) {cerr<<v[i][j]<<" ";}cerr<<endl;}


typedef long long ll;

#define vi vector<int>
#define vl vector<ll>
#define vii vector< vector<int> >
#define vll vector< vector<ll> >
#define vs vector<string>
#define pii pair<int,int>
#define pis pair<int,string>
#define psi pair<string,int>
#define pll pair<ll,ll>
template<class S, class T> pair<S, T> operator+(const pair<S, T> &s, const pair<S, T> &t) { return pair<S, T>(s.first + t.first, s.second + t.second); }
template<class S, class T> pair<S, T> operator-(const pair<S, T> &s, const pair<S, T> &t) { return pair<S, T>(s.first - t.first, s.second - t.second); }
template<class S, class T> ostream& operator<<(ostream& os, pair<S, T> p) { os << "(" << p.first << ", " << p.second << ")"; return os; }
#define X first
#define Y second
#define rep(i,n) for(int i=0;i<(n);i++)
#define rep1(i,n) for(int i=1;i<=(n);i++)
#define rrep(i,n) for(int i=(n)-1;i>=0;i--)
#define rrep1(i,n) for(int i=(n);i>0;i--)
#define REP(i,a,b) for(int i=a;i<b;i++)
#define in(x, a, b) (a <= x && x < b)
#define all(c) c.begin(),c.end()
template<class T> bool chmax(T &a, const T &b) { if (a<b) { a = b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a = b; return 1; } return 0; }
#define UNIQUE(v) v.erase(std::unique(v.begin(), v.end()), v.end());
const ll inf = 1000000001;
const ll INF = (ll)1e18 + 1;
const ll MOD = 1000000007;
//const ll MOD = 998244353;
const double pi = 3.14159265358979323846;
#define Sp(p) cout<<setprecision(15)<< fixed<<p<<endl;
int dx[4] = { -1,0, 1,0 }, dy[4] = { 0,1,0,-1 };
int dx2[8] = { 1,1,0,-1,-1,-1,0,1 }, dy2[8] = { 0,1,1,1,0,-1,-1,-1 };
#define fio() cin.tie(0); ios::sync_with_stdio(false);
// #define mp make_pair
//#define endl '\n'

ll mod_pow(ll x, ll p, ll M = MOD) {
  ll a = 1;
  while (p) {
    if (p % 2)
      a = a*x%M;
    x = x*x%M;
    p /= 2;
  }
  return a;
}

ll mod_inverse(ll a, ll M = MOD) {
  return mod_pow(a, M - 2, M);
}

int n, k;
int m = 10010;

int main() {
  cin >> n >> k;
  vll p(n + 1, vl(m));
  vector<vector<vector<pll>>> zen(n + 1, vector<vector<pll>>(m));
  zen[n][0].push_back(pll(1, 0));
  for (int i = n; i >= 0; i--) {
    rep (j, m) {
      if (zen[i][j].size() == 0) continue;
      rep (l, zen[i][j].size()) {
        (p[i][j] += zen[i][j][l].first) %= MOD;;
      }
      if (i == 0) {
        continue;
      }
      rep (x, k + 1) {
        ll nj, add;
        if (x > i) {
          nj = j;
          add = 0;
        }
        else {
          int ma = x + j;
          nj = j + ma / i;
          add = ma / i;
        }
        pll pdp;
        pdp.first = p[i][j] * mod_inverse(k + 1) % MOD;
        zen[i - 1][nj].push_back(pdp);
      }
    }
  }

  //DEBUG_MAT(p);
  //DEBUG_MAT(dp);

  ll ans = 0;
  rep (j, m) {
    // ans += p[0][j] * dp[0][j] % MOD;
    ans += p[0][j] * j % MOD;
    ans %= MOD;
  }

  ans = ans * mod_pow(k + 1, n) % MOD;
  
  ll al = 0;
  rep (x, k + 1) {
    al += x * mod_pow(k + 1, n - 1) % MOD;
    al %= MOD;
  }
  al = al * n % MOD;
  //DEBUG(al);
  ans = (al - ans + MOD) % MOD;
  cout << ans << endl;
}

### Prompt

Please formulate 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>
#include <unistd.h>

using namespace std;

#define DEBUG(x) cerr<<#x<<": "<<x<<endl;
#define DEBUG_VEC(v) cerr<<#v<<":";for(int i=0;i<v.size();i++) cerr<<" "<<v[i]; cerr<<endl
#define DEBUG_MAT(v) cerr<<#v<<endl;for(int i=0;i<v.size();i++){for(int j=0;j<v[i].size();j++) {cerr<<v[i][j]<<" ";}cerr<<endl;}


typedef long long ll;

#define vi vector<int>
#define vl vector<ll>
#define vii vector< vector<int> >
#define vll vector< vector<ll> >
#define vs vector<string>
#define pii pair<int,int>
#define pis pair<int,string>
#define psi pair<string,int>
#define pll pair<ll,ll>
template<class S, class T> pair<S, T> operator+(const pair<S, T> &s, const pair<S, T> &t) { return pair<S, T>(s.first + t.first, s.second + t.second); }
template<class S, class T> pair<S, T> operator-(const pair<S, T> &s, const pair<S, T> &t) { return pair<S, T>(s.first - t.first, s.second - t.second); }
template<class S, class T> ostream& operator<<(ostream& os, pair<S, T> p) { os << "(" << p.first << ", " << p.second << ")"; return os; }
#define X first
#define Y second
#define rep(i,n) for(int i=0;i<(n);i++)
#define rep1(i,n) for(int i=1;i<=(n);i++)
#define rrep(i,n) for(int i=(n)-1;i>=0;i--)
#define rrep1(i,n) for(int i=(n);i>0;i--)
#define REP(i,a,b) for(int i=a;i<b;i++)
#define in(x, a, b) (a <= x && x < b)
#define all(c) c.begin(),c.end()
template<class T> bool chmax(T &a, const T &b) { if (a<b) { a = b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a = b; return 1; } return 0; }
#define UNIQUE(v) v.erase(std::unique(v.begin(), v.end()), v.end());
const ll inf = 1000000001;
const ll INF = (ll)1e18 + 1;
const ll MOD = 1000000007;
//const ll MOD = 998244353;
const double pi = 3.14159265358979323846;
#define Sp(p) cout<<setprecision(15)<< fixed<<p<<endl;
int dx[4] = { -1,0, 1,0 }, dy[4] = { 0,1,0,-1 };
int dx2[8] = { 1,1,0,-1,-1,-1,0,1 }, dy2[8] = { 0,1,1,1,0,-1,-1,-1 };
#define fio() cin.tie(0); ios::sync_with_stdio(false);
// #define mp make_pair
//#define endl '\n'

ll mod_pow(ll x, ll p, ll M = MOD) {
  ll a = 1;
  while (p) {
    if (p % 2)
      a = a*x%M;
    x = x*x%M;
    p /= 2;
  }
  return a;
}

ll mod_inverse(ll a, ll M = MOD) {
  return mod_pow(a, M - 2, M);
}

int n, k;
int m = 10010;

int main() {
  cin >> n >> k;
  vll p(n + 1, vl(m));
  vector<vector<vector<pll>>> zen(n + 1, vector<vector<pll>>(m));
  zen[n][0].push_back(pll(1, 0));
  for (int i = n; i >= 0; i--) {
    rep (j, m) {
      if (zen[i][j].size() == 0) continue;
      rep (l, zen[i][j].size()) {
        (p[i][j] += zen[i][j][l].first) %= MOD;;
      }
      if (i == 0) {
        continue;
      }
      rep (x, k + 1) {
        ll nj, add;
        if (x > i) {
          nj = j;
          add = 0;
        }
        else {
          int ma = x + j;
          nj = j + ma / i;
          add = ma / i;
        }
        pll pdp;
        pdp.first = p[i][j] * mod_inverse(k + 1) % MOD;
        zen[i - 1][nj].push_back(pdp);
      }
    }
  }

  //DEBUG_MAT(p);
  //DEBUG_MAT(dp);

  ll ans = 0;
  rep (j, m) {
    // ans += p[0][j] * dp[0][j] % MOD;
    ans += p[0][j] * j % MOD;
    ans %= MOD;
  }

  ans = ans * mod_pow(k + 1, n) % MOD;
  
  ll al = 0;
  rep (x, k + 1) {
    al += x * mod_pow(k + 1, n - 1) % MOD;
    al %= MOD;
  }
  al = al * n % MOD;
  //DEBUG(al);
  ans = (al - ans + MOD) % MOD;
  cout << ans << endl;
}

```