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;
#define ll long long int
#define pb push_back
#define mp make_pair
#define nt _int128
#define mod 1000000007
ll n;
ll arr[101];
ll dp[100005][105];
ll func(ll k)
{
    for(ll i=0;i<=n;i++)
        dp[0][i]=1;
    for(ll i=0;i<=k;i++)
        dp[i][0]=1;
    for(ll i=1;i<=k;i++)
    {
        for(ll j=1;j<=n;j++)
        {
            if(arr[j]>=i)
            {
                dp[i][j]=(dp[i][j-1]+dp[i-1][j])%mod;
            }
            else
            {
                ll z=dp[i][j-1]-dp[i-arr[j]-1][j-1];
                dp[i][j]=(z+dp[i-1][j])%mod;
            }
        }
    }
  
    return dp[k][n]-dp[k-1][n];
}
int main()
{
// #ifndef ONLINE_JUDGE
//        freopen("input.txt","r",stdin);
//     freopen("output.txt","w",stdout);
// #endif
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ll t;
    // cin >> t;
    t  = 1;
    while(t--)
    {
        ll k;
        cin>>n>>k;
        for(ll i=1;i<=n;i++)
            cin>>arr[i];
        memset(dp,0,sizeof(dp));
        cout<<func(k);
    }
    return 0;
}