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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <unordered_map>
#include <unordered_set>
#include <algorithm>
#include <iostream>
#include <iostream>
#include <fstream>
#include <numeric>
#include <cstring>
#include <cassert>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <cmath>
#include <set>
#include <map>
#include <functional>
#include <bitset>
#include <iomanip>
#include <stack>
#include <list>
using namespace std;
#define     loop(n,i,a)     for(int i=a;i<n;i++)
#define     loopR(n,i,a)    for(int i=n-1;i>=a;i--)
#define     all(arr,n)      arr,arr+n
#define     allv(v)         (v).begin(),(v).end()
#define     rallv(v)        (v).rbegin(),(v).rend()
#define     m_p             make_pair
#define     ll              long long
#define     pii             pair<ll,ll>
#define     vi              vector<int>
#define     vll             vector<ll>
#define		vii 			vector<pair<ll,ll>>
#define     sz(x)           (int)x.size()
#define     pb              push_back
#define     endl            "\n"
#define     Endl            "\n"
#define     f               first
#define     s               second
#define     mem(dp,n)       memset(dp,n,sizeof dp)
#define     test(i)        cout<<i<<endl
int dx[] = { 1 , 0 ,-1 , 0 ,-1 ,-1 , 1 , 1 };
int dy[] = { 0 , 1 , 0 ,-1 ,-1 , 1 ,-1 , 1 };
void fast(){
    std::ios_base::sync_with_stdio(0);
    cin.tie(NULL);
    cout.tie(NULL);
}
const ll mxN=3e5+5,oo=0x3f3f3f3f,mod=1e9+7;
int n,k,arr[105];
ll dp[105][mxN];
ll go(int idx, int rem){
    if(rem<0)
        return 0;
    if(idx==n)
        return rem==0;
    ll &ret=dp[idx][rem];
    if(~ret)
        return ret;
    ret=((go(idx,rem-1)%mod)+(go(idx+1,rem)%mod)-(go(idx+1,rem-arr[idx]-1)))+2*mod;
    ret%=mod;
    return ret;
}
void solve(){
    cin>>n>>k;
    loop(n,i,0)
        cin>>arr[i];
    mem(dp,-1);
    cout<<go(0,k)%mod<<endl;
}
int main() {
    fast();
    solve();
}