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: Accepted)

#include <bits/stdc++.h>
#define REP(i, n) for(int i = 0; i < n; i++)
#define REPR(i, n) for(int i = n; i >= 0; i--)
#define FOR(i, m, n) for(int i = m; i < n; i++)
#define FORALL(i, m) for( auto i = m.begin(); i != m.end() ; ++i )
#define INF 2e9
#define ALL(v) v.begin(), v.end()
#define TM_T template <class T>
using namespace std;
typedef long long ll;
TM_T T inp(){T it;cin >> it;return it;}

ll N,K;
vector<ll> li;
vector<ll> dp;
vector<ll> dp2;

int input(){
  cin >> N >> K;
  REP(i,N) li.push_back( inp<ll>() );
}

int main()
{
  input();
  
  REP(i,K+1) dp.push_back(0);
  REP(i,K+1) dp2.push_back(0);
  dp[0] = 1;
  
  REP(i,N){
    REP(j,K+1) dp2[j] = 0;
    REP(j,K+1){
      dp2[j] = dp[j];
      if (j>0) dp2[j] += dp2[j-1];
      if (j-li[i]>0) dp2[j] += -dp[j-li[i]-1] + (1000000007l*9l);
    }
    REP(j,K+1) dp[j] = dp2[j]%(1000000007l);
  }
  
  cout << dp[K] << endl;
  
}