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)

#include <iostream>
#include <iomanip>
#include <cstdio>
#include <string>
#include <cstring>
#include <deque>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <algorithm>
#include <map>
#include <set>
#include <complex>
#include <cmath>
#include <limits>
#include <cfloat>
#include <climits>
#include <ctime>
#include <cassert>
#include <numeric>
#include <fstream>
#include <functional>
#include <bitset>
#define chmin(a, b) ((a)=min((a), (b)))
#define chmax(a, b) ((a)=max((a), (b)))
#define fs first
#define sc second
#define eb emplace_back
using namespace std;

typedef long long ll;
typedef pair<int, int> P;
typedef tuple<int, int, int> T;

const ll MOD=1e9+7;
const ll INF=1e18;
const double pi=acos(-1);
const double eps=1e-10;

int dx[]={1, 0, -1, 0};
int dy[]={0, -1, 0, 1};

ll dp[110][100010];
ll sum[110][100010];

int main(){
    int n, k; cin>>n>>k;
    vector<ll> a(n);
    for(int i=0; i<n; i++) cin>>a[i];

    for(int i=0; i<=a[0]; i++){
        dp[1][i] = 1;
        sum[1][i+1] = sum[1][i] + dp[1][i]; 
    }

    ll ub=a[0];
    for(int i=1; i<n; i++){
        for(ll j=0; j<=100000; j++){

            dp[i+1][j] += (sum[i][min(j+1, ub+1)] - sum[i][max(0LL, j-a[i])]);
            dp[i+1][j] += MOD;
            dp[i+1][j] %= MOD;

            sum[i+1][j+1] = sum[i+1][j] + dp[i+1][j];
            sum[i+1][j+1] %= MOD;

        }
        ub += a[i];
    }

    /*
    for(int i=0; i<=n; i++){
        cout << i << " : ";
        for(int j=0; j<=k; j++){
            cout << dp[i][j] << " ";
        }
        cout << endl;
    }
    */
    
    cout << dp[n][k] << endl;
}