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)

#include<bits/stdc++.h>
#define fo(i,a,b) for(int i= a ; i < b ; ++i)
#define rep(i,n) fo(i,0,n)
#define pln(n) printf("%lld\n",n)
#define sll(n) scanf("%lld",&n)
#define ss(n) scanf("%s",n)
#define vi vector < int >
#define pii pair < int , int >
#define pb push_back
#define mp make_pair
#define ll long long
#define ull unsigned long long
#define INF 0x3f3f3f3f
#define dbg(x) { cout<< #x << ": " << (x) << endl; }
#define dbg2(x,y) { cout<< #x << ": " << (x) << " , " << #y << ": " << (y) << endl; }
#define dbg3(x,y,z) { cout<< #x << ": " << (x) << " , " << #y << ": " << (y) <<" , "<< #z <<": "<<(z)<<endl; }
#define all(x) x.begin(),x.end()
#define sz(x) (int)x.size()
#define vll vector<long long>
#define vvi vector<vector<int>>
#define si set<int>
#define tr(c, it) \
for(decltype(c.begin()) it = c.begin(); it!= c.end(); it++)
#define pis pair<int,string>
#define present(c, x) (c.find(x) != c.end())
#define cpresent(c, x) (find(all(c),x) != c.end())
#define mod (int)(1e9 + 7) 

using namespace std;
long long dp[101][100001];

long long rec(vector<int> a, int i, int k){
	if(k == 0)return 1;
	if(i == a.size()){
		return 0;
	}
	if(dp[i][k] != -1)return dp[i][k];
	long long ret = 0;
	for(int curr = 0; curr <= a[i]; curr++){
		if(k >= curr){
			ret = ret + rec(a, i+1, k - curr);
	    	if(ret > mod)ret -= mod;
	    }	
	}
	return dp[i][k] = ret;
}

int main(){
	memset(dp, -1, sizeof(dp));
	int n, k;
	cin >> n >> k;
	vector<int> a(n);
	for(int i = 0; i < n; i++){
		cin >> a[i];
	}
	long long ret = rec(a, 0, k);
	cout << ret << endl;
	
}