#define DEB
#include<bits/stdc++.h>
#define REP(i,m) for(int i=0;i<(m);++i)
#define REPN(i,m,in) for(int i=(in);i<(m);++i)
#define ALL(t) (t).begin(),(t).end()
#define CLR(a) memset((a),0,sizeof(a))
#define pb push_back
#define mp make_pair
#define fr first
#define sc second

using namespace std;


#ifdef DEB
#define dump(x)  cerr << #x << " = " << (x) << endl
#define prl cerr<<"called:"<< __LINE__<<endl
#define dumpR(x) cerr<<"\x1b[31m"<<#x<<" = " <<(x)<<"\x1b[39m"<<endl
#define dumpY(x) cerr<<"\x1b[33m"<<#x<<" = " <<(x)<<"\x1b[39m"<<endl
#define dumpG(x) cerr<<"\x1b[32m"<<#x<<" = " <<(x)<<"\x1b[39m"<<endl
template<class T> void debug(T a,T b){ for(;a!=b;++a) cerr<<*a<<' ';cerr<<endl;}
#else
#define dump(x) ;
#define dumpR(x) ;
#define dumpY(x) ;
#define dumpG(x) ;
#define prl ;
template<class T> void debug(T a,T b){ ;}
#endif

template<class T> void chmin(T& a,const T& b) { if(a>b) a=b; }
template<class T> void chmax(T& a,const T& b) { if(a<b) a=b; }

typedef long long int lint;
typedef pair<int,int> pi;

namespace std{
  template<class S,class T>
  ostream &operator <<(ostream& out,const pair<S,T>& a){
    out<<'('<<a.fr<<','<<a.sc<<')';
    return out;
  }
}

//const int INF=5e8;

template<lint mod>
struct Int_{
  unsigned x;
  unsigned mpow(Int_ a,unsigned k){
    Int_ res=1;
    while(k){
      if(k&1) res=res*a;
      a=a*a;
      k>>=1;
    }
    return res.x;
  }
  unsigned inverse(Int_ a){
    return mpow(a,mod-2);
  }
  Int_(): x(0) { }
  Int_(long long sig) {
    int sigt=sig%mod;
    if(sigt<0) sigt+=mod;
    x=sigt;
  }
  unsigned get() const { return (unsigned)x; }
  
  Int_ &operator+=(Int_ that) { if((x += that.x) >= mod) x -= mod; return *this; }
  Int_ &operator-=(Int_ that) { if((x += mod - that.x) >= mod) x -= mod; return *this; }
  Int_ &operator*=(Int_ that) { x = (unsigned long long)x * that.x % mod; return *this; }
  Int_ &operator=(Int_ that) { x=that.x; return *this;}
  Int_ &operator/=(Int_ that) { x=(unsigned long long) x * inverse(that.x)%mod; return *this;}
  bool operator==(Int_ that) const { return x==that.x; }
  bool operator!=(Int_ that) const { return x!=that.x; }

  Int_ operator-() const { return Int_(0)-Int_(*this);}
  Int_ operator+(Int_ that) const { return Int_(*this) += that; }
  Int_ operator-(Int_ that) const { return Int_(*this) -= that; }
  Int_ operator*(Int_ that) const { return Int_(*this) *= that; }
  Int_ operator/(Int_ that) const { return Int_(*this) /= that; }
  Int_ inv(){
    return Int_(mpow(x,mod-2));
  }
};

namespace std{
  template<lint mod>
  ostream &operator <<(ostream& out,const Int_<mod>& a){
    out<<a.get();
    return out;
  }
  template<lint mod>
  istream &operator >>(istream& in,Int_<mod>& a){
    in>>a.x;
    return in;
  }
};

typedef Int_<1000000007> Int;

Int mpow(Int a,lint k){
  Int res=1;
  while(k){
    if(k&1) res=res*a;
    a=a*a;
    k>>=1;
  }
  return res;
}

const int M=105;
Int fact[M+2],invf[M+2];
void prep(){
  fact[0]=invf[0]=1;
  REP(i,M){
    fact[i+1]=fact[i]*(i+1);
  }
  invf[M]=fact[M].inv();
  for(int i=M-1;i>=0;--i) invf[i]=invf[i+1]*(i+1);
}
Int C(int a,int b){
  if(a<0 || b<0 || a<b) return 0;
  return fact[a]*invf[a-b]*invf[b];
}

int n,K;

Int dp[105][105*105];//n.o. pattern
Int dp2[105][105*105];//sum of left ishi

Int expK[105];
int main(){
  cin>>n>>K;

  dp[0][0]=1;
  dp2[0][0]=0;
  int m=n*K;
  REP(i,n){
    for(int j=0;j<=m;++j) if(dp[i][j].x || dp2[i][j].x){
      REP(k,K+1){
        if(k<=n-i){
          int incr=(k+j)/(n-i);
          dp[i+1][j+incr]+=dp[i][j];
          dp2[i+1][j+incr]+=dp2[i][j]+dp[i][j]*((k+j)%(n-i));
        }else{
          dp[i+1][j]+=dp[i][j];
          dp2[i+1][j]+=dp2[i][j]+dp[i][j]*(k+j);
        }
      }
    }
  }
  Int res=0;
  REP(i,m+1) res+=dp2[n][i];
  cout<<res<<endl;
  return 0;
}




### Prompt

Please create a solution in cpp 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
#define DEB
#include<bits/stdc++.h>
#define REP(i,m) for(int i=0;i<(m);++i)
#define REPN(i,m,in) for(int i=(in);i<(m);++i)
#define ALL(t) (t).begin(),(t).end()
#define CLR(a) memset((a),0,sizeof(a))
#define pb push_back
#define mp make_pair
#define fr first
#define sc second

using namespace std;


#ifdef DEB
#define dump(x)  cerr << #x << " = " << (x) << endl
#define prl cerr<<"called:"<< __LINE__<<endl
#define dumpR(x) cerr<<"\x1b[31m"<<#x<<" = " <<(x)<<"\x1b[39m"<<endl
#define dumpY(x) cerr<<"\x1b[33m"<<#x<<" = " <<(x)<<"\x1b[39m"<<endl
#define dumpG(x) cerr<<"\x1b[32m"<<#x<<" = " <<(x)<<"\x1b[39m"<<endl
template<class T> void debug(T a,T b){ for(;a!=b;++a) cerr<<*a<<' ';cerr<<endl;}
#else
#define dump(x) ;
#define dumpR(x) ;
#define dumpY(x) ;
#define dumpG(x) ;
#define prl ;
template<class T> void debug(T a,T b){ ;}
#endif

template<class T> void chmin(T& a,const T& b) { if(a>b) a=b; }
template<class T> void chmax(T& a,const T& b) { if(a<b) a=b; }

typedef long long int lint;
typedef pair<int,int> pi;

namespace std{
  template<class S,class T>
  ostream &operator <<(ostream& out,const pair<S,T>& a){
    out<<'('<<a.fr<<','<<a.sc<<')';
    return out;
  }
}

//const int INF=5e8;

template<lint mod>
struct Int_{
  unsigned x;
  unsigned mpow(Int_ a,unsigned k){
    Int_ res=1;
    while(k){
      if(k&1) res=res*a;
      a=a*a;
      k>>=1;
    }
    return res.x;
  }
  unsigned inverse(Int_ a){
    return mpow(a,mod-2);
  }
  Int_(): x(0) { }
  Int_(long long sig) {
    int sigt=sig%mod;
    if(sigt<0) sigt+=mod;
    x=sigt;
  }
  unsigned get() const { return (unsigned)x; }
  
  Int_ &operator+=(Int_ that) { if((x += that.x) >= mod) x -= mod; return *this; }
  Int_ &operator-=(Int_ that) { if((x += mod - that.x) >= mod) x -= mod; return *this; }
  Int_ &operator*=(Int_ that) { x = (unsigned long long)x * that.x % mod; return *this; }
  Int_ &operator=(Int_ that) { x=that.x; return *this;}
  Int_ &operator/=(Int_ that) { x=(unsigned long long) x * inverse(that.x)%mod; return *this;}
  bool operator==(Int_ that) const { return x==that.x; }
  bool operator!=(Int_ that) const { return x!=that.x; }

  Int_ operator-() const { return Int_(0)-Int_(*this);}
  Int_ operator+(Int_ that) const { return Int_(*this) += that; }
  Int_ operator-(Int_ that) const { return Int_(*this) -= that; }
  Int_ operator*(Int_ that) const { return Int_(*this) *= that; }
  Int_ operator/(Int_ that) const { return Int_(*this) /= that; }
  Int_ inv(){
    return Int_(mpow(x,mod-2));
  }
};

namespace std{
  template<lint mod>
  ostream &operator <<(ostream& out,const Int_<mod>& a){
    out<<a.get();
    return out;
  }
  template<lint mod>
  istream &operator >>(istream& in,Int_<mod>& a){
    in>>a.x;
    return in;
  }
};

typedef Int_<1000000007> Int;

Int mpow(Int a,lint k){
  Int res=1;
  while(k){
    if(k&1) res=res*a;
    a=a*a;
    k>>=1;
  }
  return res;
}

const int M=105;
Int fact[M+2],invf[M+2];
void prep(){
  fact[0]=invf[0]=1;
  REP(i,M){
    fact[i+1]=fact[i]*(i+1);
  }
  invf[M]=fact[M].inv();
  for(int i=M-1;i>=0;--i) invf[i]=invf[i+1]*(i+1);
}
Int C(int a,int b){
  if(a<0 || b<0 || a<b) return 0;
  return fact[a]*invf[a-b]*invf[b];
}

int n,K;

Int dp[105][105*105];//n.o. pattern
Int dp2[105][105*105];//sum of left ishi

Int expK[105];
int main(){
  cin>>n>>K;

  dp[0][0]=1;
  dp2[0][0]=0;
  int m=n*K;
  REP(i,n){
    for(int j=0;j<=m;++j) if(dp[i][j].x || dp2[i][j].x){
      REP(k,K+1){
        if(k<=n-i){
          int incr=(k+j)/(n-i);
          dp[i+1][j+incr]+=dp[i][j];
          dp2[i+1][j+incr]+=dp2[i][j]+dp[i][j]*((k+j)%(n-i));
        }else{
          dp[i+1][j]+=dp[i][j];
          dp2[i+1][j]+=dp2[i][j]+dp[i][j]*(k+j);
        }
      }
    }
  }
  Int res=0;
  REP(i,m+1) res+=dp2[n][i];
  cout<<res<<endl;
  return 0;
}




```