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: Runtime Error)

//#pragma GCC optimize "trapv"
#include<bits/stdc++.h>
#define ll long long int
#define fab(a,b,i) for(int i=a;i<b;i++)
#define pb push_back
#define db double
#define mp make_pair
#define endl "\n"
#define f first
#define se second
#define all(x) x.begin(),x.end()
#define MOD 1000000007
#define quick ios_base::sync_with_stdio(false);cin.tie(NULL)
using namespace std;
ll dp[105][100005],n,k,a[105];

int main()
{ quick;

	cin>>n>>k;
	fab(0,n,i)
	cin>>a[i];
	ll pref[k+1]={0};
	dp[0][0]=1;
	pref[0]=1;
	fab(1,a[0]+1,i)
	{
		dp[0][i]=1;
		//pref[i]=pref[i-1]+dp[0][i];

	}
	fab(1,k+1,i)
	pref[i]=(pref[i-1]+dp[0][i])%MOD;
	fab(0,n,i)
	dp[i][0]=1;

	fab(1,n,i)
	
	{
		fab(1,k+1,j)
		{
			ll tot=max((ll)(0),j-a[i]),op;
			if(tot==0)
				op=(pref[j])%MOD;
			else
				op=(pref[j]-pref[tot-1])%MOD;
			assert(op>=0);
			op=op+MOD;

			dp[i][j]=(op)%MOD;//+(tot==0?j:a[i]);


			

		}
		//cout<<pref[0]<<" ";
		fab(1,k+1,j)
		{
			//cout<<pref[j]<<" ";
			pref[j]=(pref[j-1]%MOD+dp[i][j])%MOD;
		}
		//cout<<endl;

	}
	//cout<<"\n\n";
	/*fab(0,n,i)
	{fab(0,k+1,j)
	cout<<dp[i][j]<<" ";
	cout<<endl;}*/
	cout<<dp[n-1][k]%MOD;


	return 0;
}