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>
#define REP(i,n) for(int i=0,i##_len=(n);i<i##_len;++i)
#define int long long
using namespace std;
typedef pair<int,int> P;
constexpr int mod=1e9+7;
int N,K;
int a[100];
int dp[101][200002]={};

signed main(){
    cin>>N>>K;
    REP(i,N) cin>>a[i];
    dp[N][K]=1;
    for(int i=N-1;i>=0;--i) for(int j=K;j>=0;--j){
        dp[i][j]=dp[i+1][j]+dp[i][j+1]-dp[i+1][j+a[i]+1];
        dp[i][j]=(dp[i][j]+mod)%mod;
    }
    cout<<dp[0][0]<<endl;
}