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: Runtime Error)

#include<iostream>
#include<vector>
#include<iomanip>
#include<algorithm>
#include<string>

using namespace std;
 
typedef long long ll;

ll memo[110][101000];

const ll MOD = 1e9 + 7;

ll a[110];

ll dp(ll i, ll j){
  if( j == 0){ return memo[i][j] = 1;}
  if( j < 0){ return 0;} //インデックスがマイナスのやばい場合を除外
  if( i < 0){ return 0;}
  //初期値 i == 0(0人に配る）も一通りとして考える
  if( memo[i][j] != -1){ return memo[i][j];}
  //すでに書き込まれている場合
  return memo[i][j] = ( (dp(i-1,j) + dp(i,j-1)) % MOD - dp( i-1,j-a[i] - 1)  +  MOD )% MOD;}
                       
  int main(){
    ll N,K; cin >> N >> K;
    for(ll i = 0; i < 110; i++){
      for(ll j = 0; j < 101000; j++){
        memo[i][j] = -1;}} //初期化(未記入)
    
    for(ll i = 0; i < N; i++){
      cin >> a[i];} //i番目の個の上限記入
    
    cout << dp(N-1,K) <<endl; return 0;}