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++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define int long long
#define pb push_back
#define myPair pair<int,int>
#define mp make_pair
#define fi first
#define se second
#define qmax priority_queue<int>
#define qmin priority_queue<int,vector<int>,greater<int>>
#define all(x) x.begin(),x.end()
#define FOR(i,a,b) for(int i = a; i <= b; i++)
#define FORR(i,a,b) for(int i = a; i >= b; i--)

const int mxn = 1e5 + 10, mod = 1e9 + 7;
int dp[101][mxn], aux[101][mxn];
int n,k;
int a[mxn];

int32_t main()
{
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif

	ios_base::sync_with_stdio(false);
	cin.tie(NULL);

	cin >> n >> k;
	FOR(i,1,n)
		cin >> a[i];
	FOR(j,1,k)
	{
		dp[0][j] = 0;
		aux[0][j] = 1;
	}
	aux[0][0] = dp[0][0] = 1;

	FOR(i,1,n)
	{
		aux[i][0] = dp[i][0] = 1;
		FOR(j,1,k)
		{
			int count = j - min(j, a[i]);
			if(count == 0)
				dp[i][j] = aux[i - 1][j];
			else
				dp[i][j] = (aux[i - 1][j] - aux[i - 1][count - 1] + mod) % mod;
		}
		FOR(j,1,k)
			aux[i][j] = (aux[i][j - 1] + dp[i][j]) % mod;
	}
	cout << dp[n][k];

	return 0;
}