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;
#define ll long long
#define mp make_pair
#define pb push_back
#define scanVec(vec , n) for(int i = 0; i < n ; i++){ cin>>vec[i];}
#define printVec(vec , n) for(int i = 0; i < n ; i++){ cout<<vec[i]<<" ";}
#define S second
#define F first
const int  MOD = 1e9 + 7;
const int  N = 1e5 + 7;
ll dp[102][N] , prefix[102][N] , a[102];

inline int add(int a, int b){
	a += b;
	if(a >= MOD)a -= MOD;
	return a;
}
 
inline int sub(int a, int b){
	a -= b;
	if(a < 0)a += MOD;
	return a;
}

//  a d f j k  l  ;
int main(){
    ios::sync_with_stdio(0);  cin.tie(0);
    ll n , k;
    cin>>n>>k;
    for(ll i = 1 ; i <= n ; i++) cin>>a[i];
    
    for(ll i = 0;i <= n; i++) dp[i][0] = prefix[i][0] = 1;
	for(ll j = 0;j <= k; j++) prefix[0][j] = 1;

    for(ll i = 1 ; i <= n ; i++){
        for(ll j = 1 ; j <= k ; j++){
            int first = (j - a[i] - 1 >= 0)? prefix[i - 1][j - a[i] - 1] : 0;
			int last = prefix[i - 1][j];
			dp[i][j] = add(dp[i][j], sub(last, first));
			prefix[i][j] = add(prefix[i][j - 1], dp[i][j]);
        }
    }
    cout<<dp[n][k]<<endl;

    return 0;
}