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: Wrong Answer)


//   //  /////       ///      
//  //   //  //     // //     
// //    //   //   //   //    
////     //   //  //     //   
// //    //   //  /////////    
//  //   //  //   //     //    
//   //  /////    //     //    

#include <bits/stdc++.h>
using namespace std;
#define forstl(i,v) for(auto &i: v)
#define forn(i,e) for(int i = 0; i < e; i++)
#define forsn(i,s,e) for(int i = s; i < e; i++)
#define rforn(i,s) for(int i = s; i >= 0; i--)
#define rforsn(i,s,e) for(int i = s; i >= e; i--)
#define iter(it,a,b) for(a::iterator it=b.begin(); it!=b.end();++it)
#define clz(a) __builtin_clz(a) // count leading zeroes
#define ctz(a) __builtin_ctz(a) // count trailing zeroes
#define popc(a) __builtin_popcount(a) // count set bits (binary me 1 kitne hai) (if longlong popcount ll)
#define ln "\n"
#define dbg(x) cerr<<#x<<" = "<<x<<endl
#define dbg2(x,y) cerr<<#x<<" = "<<x<<" & "<<#y<<" = "<<y<<endl;
#define dbgstl32(v) cerr<<#v<<" = \n"; { int c=0; forstl(it,v) \
cerr<<"    Term "<< ++c <<" = "<<it<<ln;} cerr<<endl
#define dbgstlp32(v) cerr<<#v<<" = \n"; { int c=0; forstl(it,v) \
cerr<<"    Term "<< ++c <<" = "<<it.fi<<" , "<<it.se<<ln;} cerr<<endl
#define dbgarr(v,s,e) cerr<<#v<<" = "; forsn(i,s,e) cerr<<v[i]<<", "; cerr<<endl
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define all(n) n.begin() , n.end() 
#define tfi first.first
#define tse first.second
#define tth second
typedef long long ll;
typedef long double ld;
typedef pair<int,int> p32; 
typedef pair<ll,ll> p64;
typedef pair<double,double> pdd;
typedef vector<ll> v64;
typedef vector<int> v32;
typedef vector<vector<int> > vv32;
typedef vector<p64> vp64;
typedef vector<p32> vp32;
typedef map<int,int> m32;
typedef pair< pair<int,int> , int> tri;
typedef vector<bool> vbool;
typedef priority_queue<p32> pq32;
typedef priority_queue<p32, vector<p32>, greater<p32> > pq32l;

#pragma GCC optimize ("-O2")
const ll LIM = 2e5+5;
const ll MOD = 1e9+7;
const ll inf = 1e10;



int main(){
	ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);

	int n, k;
	cin >> n >> k;

	ll a[n];
	forn(i,n) cin >> a[i];

	ll dp[n+1][k+1];

	forn(i,k+1) dp[0][i] = 0;
	forn(i,n+1) dp[i][0] = 1;

	forn(i,n){
		forn(j,k){
			if(j >= a[i]) dp[i+1][j+1] = (dp[i][j+1] + dp[i+1][j] - dp[i][j-a[i]])%MOD;
			else dp[i+1][j+1] = (dp[i][j+1] + dp[i+1][j])%MOD;
		}
	}

	cout << setprecision(20) << dp[n][k];


}