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 ll long long int
#define ull unsigned long long int
#define ld long double
#define pb push_back
#define pf pop_front
#define mp make_pair
#define mod 1000000007
using namespace std;
#define int ll

signed main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	
	int n,k;
	cin>>n>>k;
	int a[n];
	for(int i=0;i<n;i++)
		cin>>a[i];
	int dp[k+1][n+1];
	for(int i=0;i<=k;i++)
		for(int j=0;j<=n;j++)
			dp[i][j]=0;
	
	for(int i=0;i<=k;i++)
		dp[i][0]=0;
	for(int i=0;i<=min(k,a[0]);i++)
		dp[i][1]=1;
	for(int i=2;i<=n;i++)
	{
		dp[0][i]=1;
		int su=1;
		for(int j=1;j<=k;j++)
		{
			su=(su+dp[j][i-1])%mod;
			if(j>a[i-1])
				su=((su-dp[j-a[i-1]-1][i-1])%mod+mod)%mod;
			dp[j][i]=su;
			if(su==1)
				break;
			//cout<<dp[j][i]<<' ';
		}
		//cout<<endl;
	}
	//for(int i=0;i<=n;i++)
	//{
	//	for(int j=0;j<=k;j++)
	//		cout<<dp[j][i]<<' ';
	//	cout<<endl;
	//}
	cout<<(dp[k][n]%mod+mod)%mod<<endl;
}