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;
typedef long long int lli;
typedef vector <int> vi;
typedef vector <vi> vvi;
typedef pair <int,int> pr;
#define fr(i, n) for(int i = 0; i < n; i++)
#define sz(a) int((a).size())
#define pb push_back
#define all(c)(c).begin(), (c).end()
#define tr(c, i) for (typeof ((c).begin() i = (c).begin(); i != (c).end(); i++)
#define present(c, x)((c).find(x) != (c).end())
#define cpresent(c, x)(find(all(c), x) != (c).end())
// for mod power a^b%m use power<lli> (a,b,m)
template < class mytype >
mytype power(mytype a, mytype b, mytype m){if (b == 1) return a % m;if (b == 0) return 1;
mytype root = power < lli > (a, b / 2, m);mytype ans = (root * root) % m;if (b % 2) ans = (ans * a) % m;
return ans;
}
// for gcd use gcd<int> (a,b)
template < class type >type gcd(type a, type b){if (a == 0) return b;else return gcd(b % a, a);}
template < class type >type mod_inv(type a, type p) {return power(a, p - 2, p);}
const lli MOD = 1e9 + 7;

int main()
{
   int n, k;
   cin >> n >> k;
   vector<int> v(n);
   for(int i = 0 ; i < n ; i++ ) cin >> v[i];
    lli dp[k+1];
   for(int i = 0 ; i <= k ; i++) {
        if(i <= v[0]) dp[i] = 1;
        else dp[i] = 0;
   }lli sum = v[0];
   for(int i = 1 ; i < n ; i++){
    sum += v[i];int count = 0;
    for(int j = 1 ; j <= k && j < sum; j++) {
            if(j <=  v[i])dp[j] = (dp[j] + dp[j-1]) % MOD;
            else{
                dp[j] = (dp[j] + dp[j-1] - dp[count]) % MOD;
                count++;
            }
    }

    if(sum <= k) dp[sum] = 1;
   }
   cout << dp[k] << endl;
    return 0;
}