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: Time Limit Exceeded)

#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/detail/standard_policies.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef tree<pair<int,int>, null_type, less<pair<int,int>>, rb_tree_tag,
		tree_order_statistics_node_update> ordered_set;

long long mem[100005][105];
bool vis[100005][105];
long long arr[105];
int n , k;
const long mod = 1e9 + 7;
long long sol(long long rem , int idx){
	if(rem==0)
		return 1;
	if(rem < 0 || idx == n)
		return 0;
	if(vis[rem][idx])
		return mem[rem][idx];
	long long ret = (((sol(rem-1,idx)%mod)+(sol(rem,idx+1)%mod) - sol(rem-arr[idx]-1,idx+1)%mod)+mod)%mod;
	return mem[rem][idx] = ret;
}
int main(){
	cin>>n>>k;
	for(int i = 0; i < n ; i++)
		cin>>arr[i];
	cout<<sol(k,0);
	return 0;
}