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 _                             \
    ios_base::sync_with_stdio(false); \
    cin.tie(NULL);                    \
    cout.tie(NULL);
#define deb(x) cerr << "\n" \
                    << #x << "=" << x << "\n";
#define deb2(x, y) cerr << "\n"                   \
                        << #x << "=" << x << "\n" \
                        << #y << "=" << y << "\n";
typedef long long int ll;

const int mod = 1e9+7;

int main() {
	int n,k;
	cin>>n>>k;
	
	int a[n+1];
	
	for(int i=0;i<n;i++)
		cin>>a[i];
		
	int dp[n+1][k+1];
	
	memset(dp,0,sizeof(dp));
	
	
	for(int i=0;i<n;i++)
	{
		dp[i][0] = 1;
	}
	
	for(int i=1;i<=k;i++)
	{
		if(a[0]>=i)
			dp[0][i]=(dp[0][i-1]+1)%mod;
		else
			dp[0][i]=dp[0][i-1];
	}
	if(n==1)
	{
		cout<<dp[0][k]-dp[0][k-1];
		return 0;
	}
	
	for(int cand= 1;cand<=k;cand++)
	{
		for(int i=1;i<n-1;i++)
		{
			int temp = 0;
	
			if(cand>a[i])
				temp = (dp[i-1][cand]%mod - dp[i-1][cand-a[i]-1]%mod + mod )%mod;
			else
				temp = dp[i-1][cand]%mod;
			dp[i][cand] = (dp[i][cand-1]%mod+temp%mod)%mod;
		}
	}
	
	
	if(k>a[n-1])
		cout<<(dp[n-2][k]%mod-dp[n-2][k-a[n-1]-1]%mod + mod)%mod;
	else
		cout<<dp[n-2][k];
	
    return 0;
}