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  MOD 1000000007

using namespace std;
typedef long long ll;

int n,arr[105],k;
ll dp[105][100005];

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);cout.tie(nullptr);
    cin>>n>>k;
    for(int i=1;i<=n;i++)cin>>arr[i];
    for(int i=0;i<=n;i++)dp[i][0]=1;
    for(int i=0;i<=k;i++)dp[0][i]=1;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=k;j++){
            ll sum=dp[i-1][j];
            if(j-arr[i]-1>=0)sum-=dp[i-1][j-arr[i]-1];
            while(sum<0)sum+=MOD;
            dp[i][j]=(dp[i][j-1]+sum)%MOD;
        }
    }
    /*for(int i=0;i<=n;i++){
        for(int j=0;j<=k;j++){
            cout<<dp[i][j]<<' ';
        }
        cout<<endl;
    }*/
    /*if(k==0)cout<<1;
    else*/
    ll res=dp[n][k]-dp[n][k-1];
    while(res<0)res+=MOD;
    cout<<res;
}