Problem:
Score : 100 points

Problem Statement

There are N children, numbered 1, 2, \ldots, N.

They have decided to share K candies among themselves.
Here, for each i (1 \leq i \leq N), Child i must receive between 0 and a_i candies (inclusive).
Also, no candies should be left over.

Find the number of ways for them to share candies, modulo 10^9 + 7.
Here, two ways are said to be different when there exists a child who receives a different number of candies.

Constraints

All values in input are integers.

1 \leq N \leq 100

0 \leq K \leq 10^5

0 \leq a_i \leq K

Input

Input is given from Standard Input in the following format:

N K
a_1 a_2 \ldots a_N

Output

Print the number of ways for the children to share candies, modulo 10^9 + 7.

Sample Input 1

3 4
1 2 3

Sample Output 1

5

There are five ways for the children to share candies, as follows:

(0, 1, 3)

(0, 2, 2)

(1, 0, 3)

(1, 1, 2)

(1, 2, 1)

Here, in each sequence, the i-th element represents the number of candies that Child i receives.

Sample Input 2

1 10
9

Sample Output 2

0

There may be no ways for the children to share candies.

Sample Input 3

2 0
0 0

Sample Output 3

1

There is one way for the children to share candies, as follows:

(0, 0)

Sample Input 4

4 100000
100000 100000 100000 100000

Sample Output 4

665683269

Be sure to print the answer modulo 10^9 + 7.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
using Int = long long;
template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}


template<typename T,T MOD = 1000000007>
struct Mint{
  T v;
  Mint():v(0){}
  Mint(signed v):v(v){}
  Mint(long long t){v=t%MOD;if(v<0) v+=MOD;}

  Mint pow(long long k){
    Mint res(1),tmp(v);
    while(k){
      if(k&1) res*=tmp;
      tmp*=tmp;
      k>>=1;
    }
    return res;
  }
  
  Mint inv(){return pow(MOD-2);}
  
  Mint& operator+=(Mint a){v+=a.v;if(v>=MOD)v-=MOD;return *this;}
  Mint& operator-=(Mint a){v+=MOD-a.v;if(v>=MOD)v-=MOD;return *this;}
  Mint& operator*=(Mint a){v=1LL*v*a.v%MOD;return *this;}
  Mint& operator/=(Mint a){return (*this)*=a.inv();}
  
  Mint operator+(Mint a) const{return Mint(v)+=a;};
  Mint operator-(Mint a) const{return Mint(v)-=a;};
  Mint operator*(Mint a) const{return Mint(v)*=a;};
  Mint operator/(Mint a) const{return Mint(v)/=a;};

  Mint operator-(){return v?MOD-v:v;}

  bool operator==(const Mint a)const{return v==a.v;}
  bool operator!=(const Mint a)const{return v!=a.v;}
  bool operator <(const Mint a)const{return v <a.v;}

  // find x s.t. a^x = b
  static T log(Mint a,Mint b){
    const T sq=40000;
    unordered_map<T, T> dp;
    dp.reserve(sq);
    Mint res(1);
    for(int r=0;r<sq;r++){
      if(!dp.count(res)) dp[res]=r;
      res*=a;
    }
    Mint p=pow(a.inv(),sq);
    res=b;
    for(int q=0;q<=MOD/sq+1;q++){
      if(dp.count(res)){
        T idx=q*sq+dp[res];
        if(idx>0) return idx;
      }
      res*=p;
    }    
    return T(-1);
  }

  static vector<Mint> fact,finv,invs;
  
  static void init(int n){
    if(n+1<=(signed)fact.size()) return;
    fact.assign(n+1,1);
    finv.assign(n+1,1);
    invs.assign(n+1,1);
    
    for(int i=1;i<=n;i++) fact[i]=fact[i-1]*Mint(i);
    finv[n]=Mint(1)/fact[n];
    for(int i=n;i>=1;i--) finv[i-1]=finv[i]*Mint(i);
    for(int i=1;i<=n;i++) invs[i]=finv[i]*fact[i-1];
  }

  static Mint comb(long long n,int k){
    Mint res(1);
    for(int i=0;i<k;i++){
      res*=Mint(n-i);
      res/=Mint(i+1);
    }
    return res;
  }
  
  static Mint C(int n,int k){
    if(n<k||k<0) return Mint(0);
    init(n);
    return fact[n]*finv[n-k]*finv[k];
  }

  static Mint P(int n,int k){
    if(n<k||k<0) return Mint(0);
    init(n);
    return fact[n]*finv[n-k];
  }
  
  static Mint H(int n,int k){
    if(n<0||k<0) return Mint(0);
    if(!n&&!k) return Mint(1);
    init(n+k-1);
    return C(n+k-1,k);
  }

  static Mint S(int n,int k){
    Mint res;
    init(k);
    for(int i=1;i<=k;i++){
      Mint tmp=C(k,i)*Mint(i).pow(n);
      if((k-i)&1) res-=tmp;
      else res+=tmp;
    }    
    return res*=finv[k];
  }

  static vector<vector<Mint> > D(int n,int m){
    vector<vector<Mint> > dp(n+1,vector<Mint>(m+1,0));
    dp[0][0]=Mint(1);
    for(int i=0;i<=n;i++){
      for(int j=1;j<=m;j++){
        if(i-j>=0) dp[i][j]=dp[i][j-1]+dp[i-j][j];
        else dp[i][j]=dp[i][j-1];
      }
    }
    return dp;
  }

  static Mint B(int n,int k){
    Mint res;
    for(int j=1;j<=k;j++) res+=S(n,j);
    return res;
  }

  static Mint montmort(int n){
    Mint res;
    init(n);
    for(int k=2;k<=n;k++){
      if(k&1) res-=finv[k];
      else res+=finv[k];
    }
    return res*=fact[n];
  }

  static Mint LagrangePolynomial(vector<Mint> &y,Mint t){
    int n=y.size()-1;    
    if(t.v<=n) return y[t.v];
    init(n+1);
    Mint num(1);
    for(int i=0;i<=n;i++) num*=t-Mint(i);
    Mint res;
    for(int i=0;i<=n;i++){
      Mint tmp=y[i]*num/(t-Mint(i))*finv[i]*finv[n-i];
      if((n-i)&1) res-=tmp;
      else res+=tmp;
    }
    return res;
  }
};
template<typename T,T MOD>
vector<Mint<T, MOD> > Mint<T, MOD>::fact = vector<Mint<T, MOD> >();
template<typename T,T MOD>
vector<Mint<T, MOD> > Mint<T, MOD>::finv = vector<Mint<T, MOD> >();
template<typename T,T MOD>
vector<Mint<T, MOD> > Mint<T, MOD>::invs = vector<Mint<T, MOD> >();

//INSERT ABOVE HERE
signed main(){
  int n,k;
  cin>>n>>k;
  vector<int> a(n);
  for(int i=0;i<n;i++) cin>>a[i];

  using M = Mint<int>;
  vector<M> dp(k+1,0);
  dp[0]=M(1);
  for(int x:a){
    vector<M> nx(k+1,0);
    M tmp;
    for(int i=0;i<=k;i++){
      tmp+=dp[i];
      if(i>x) tmp-=dp[i-(x+1)];
      nx[i]=tmp;
    }
    swap(dp,nx);
  }
  cout<<dp[k].v<<endl;
  return 0;
}