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)

// We only fail when we stop trying
#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
#define endl '\n'
#define D(x) cerr << #x << " = " << (x) << '\n'
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(), (x).end()
typedef long long ll;

const int N = 105;
const int V = 1e5 + 5;
const int Mod = 1e9 + 7;

int n, k;
int a[N];
int dp[V];
int acc[V];

int main()
{
  cin >> n >> k;
  for(int i = 0 ; i < n ; i++) {
    cin >> a[i];
  }
  for(int j = 0; j <= k; j++)
    acc[j] = 1;
  for(int i = 0 ; i < n ; i++) {
    for(int j = 0 ; j <= k; j++) {
      dp[j] = acc[j];
      if(j - a[i] - 1 >= 0)
        dp[j] = ((dp[j] - acc[j - a[i] - 1]) % Mod + Mod) % Mod;
    }
    acc[0] = dp[0];
    for(int j = 1 ; j <= k ; j++)
      acc[j] = (acc[j - 1] + dp[j]) % Mod;
  }
  cout << dp[k] << endl;
  return 0;
}