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++ (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <bits/stdc++.h>
using namespace std;

typedef vector<int> vi;
#define ll long long
#define PI 3.14159265359
//#define x first
//#define y second
#define point pair<int,int>
#define clr(x, val)	memset((x), (val), sizeof(x))

//you don't know math, unless you can do math!
int mod = 1e9 + 7;
int dp[109][100009];
int main(){
    cin.tie(0);
    cin.sync_with_stdio(0);

    int n,k;
    cin>>n>>k;

    vector<int> a(n);
    for(int i = 0; i < n; i++)
        cin>>a[i];



    for(int i = 0; i <= n; i++)
        dp[i][0]= 1;

    for(int i = n - 1; i >= 0; i--){
        int sums[k + 1];
        sums[0] = dp[i + 1][0];
        for(int j = 1; j <= k; j++)
            sums[j] = (sums[j - 1] + dp[i + 1][j])%mod;

        for(int j = 0; j <= k; j++){
            int sum  = sums[j];
            if(j - a[i] - 1 >= 0) sum = (sum - sums[j - a[i] - 1] + mod)%mod;

            dp[i][j] = sum;
            //dp[i][j] = dp[i + 1][j - x : 0 <= x <= a[i]]
        }
    }

    cout<<dp[0][k];
}