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: Time Limit Exceeded)

//apigs property
//Happiness can be found, even in the darkest of times, if one only remembers to turn on the light
#include<bits/stdc++.h>
#define fast ios_base::sync_with_stdio(0) , cin.tie(0) , cout.tie(0)
#define F first
#define S second
#define pb push_back
#define vll vector< ll >
#define vi vector< int >
#define pll pair< ll , ll >
#define pi pair< int , int >
#define all(s) s.begin() , s.end()
#define sz(s) s.size()
#define md (s + e) / 2
#define mid (l + r) / 2
#define msdp(dp) memset(dp , -1 , sizeof dp)
#define mscl(dp) memset(dp , 0 , sizeof dp)
#define C continue
#define R return
#define lx node * 2
#define rx node * 2 + 1
using namespace std;
typedef long long ll;
ll q, d, dp[100][100005], a[555555] , b[555555], k, l, m, n, o, p;
map < ll , ll > mp;
vll adj[555555];
const ll mod = 1e9+7;
ll mem(ll i , ll sum){
	if(sum > k)R 0;
	if(i == n)R sum == k;
	ll &r = dp[i][sum];
	if(r != -1)R r;
	r = 0;
	for(ll j = 0 ; j <= a[i] ; j++){
		r += mem(i + 1 , sum + j) % mod;
	}
	r %= mod;
	R r;
}
void solve(){
	msdp(dp);
	cin >> n >> k;
	for(ll i = 0 ; i < n ; i++){
		cin >> a[i];
	}
	cout << mem(0 , 0);
}
int main(){
    fast ;
	//cin >> q;
    q = 1;
    while(q--){
    	solve();
	}
}