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>
#define REP(i, n) for (int i = 0; i < (int)n; i++)
#define REPvec(itr, mp) for (auto itr = mp.begin(); itr != mp.end(); itr++)
#define all(x) x.begin(), x.end()
#define MOD 1000000007  // 1e9+7
using namespace std;
typedef long long ll;
typedef vector<bool> vecbool;
typedef vector<int> vecint;
typedef vector<ll> vecll;
typedef vector<string> vecstr;
typedef vector<vecint> vec2int;
typedef vector<vecbool> vec2bool;
int dx[4]{-1, +1, 0, 0};
int dy[4]{0, 0, -1, +1};
void init() {
  cin.tie(nullptr);
  ios_base::sync_with_stdio(false);
  cout << fixed << setprecision(15);
}
int main() {
  init();
  int n, k;
  cin >> n >> k;
  vecint a(n);
  for (auto& e : a) cin >> e;
  vec2int dp(n + 1, vecint(k + 1, 0));
  REP(i, n + 1) dp[i][0] = 1;
  REP(i, n) {
    REP(j, k) {
      dp[i + 1][j + 1] += dp[i][j + 1] + dp[i + 1][j];
      if (j - a[i] >= 0) {
        dp[i + 1][j + 1] -= dp[i + 1][j - a[i]];
      }
      dp[i + 1][j + 1] %= MOD;
    }
  }
  cout << dp[n][k] << endl;
}