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)

#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize ("Ofast")
#pragma GCC target ("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
 
 
#include <iostream>
#include <iomanip>
#include <sstream>
#include <cstring>
#include <vector>
#include <deque>
#include <queue>
#include <set>
#include <map>
#include <valarray>
#include <iterator>
#include <functional>
#include <limits>
#include <algorithm>
#include <numeric>
#include <cmath>
#include <cassert>
#include <unordered_map>
#include <unordered_set>
#include <stack>
using namespace std;
#define IOS ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
 
typedef int64_t ll;
 
#define endl '\n'
#define int ll
ll mod=1e9+7;
ll mod1=1e9+5;
ll power(ll a,ll b)
{
    if(b==0) return 1;
    else if(b%2==0)
        return power((((a%mod)*(a%mod))%mod),b/2)%mod;
    else return ((a%mod)*(power((((a%mod)*(a%mod))%mod),b/2)%mod))%mod;
}


int dp[100001][101];
int cum[100001][101];
int32_t main()
{
    IOS
    
    
    int n,k;
    cin>>n>>k;
    int a[n];
    for(int i=0;i<n;++i)
        cin>>a[i];
    
    
    memset(dp,0,sizeof(dp));
    memset(cum,0,sizeof(cum));
    for(int i=1 ;i<=n;++i)
        dp[0][i]=1,cum[0][i]=1;
    
    
   for(int i=1;i<=n;++i)
   {
        dp[1][i]=dp[1][i-1];
        if(a[i-1]>0)
            dp[1][i]++;
        cum[1][i]=(cum[0][i]+dp[1][i])%mod;
   }
   for(int i=1;i<=k;++i)
   {
        cum[i][1]=cum[i-1][1];
        if(i<=a[0])
        {
            cum[i][1]++;
        }
        cum[i][1]%=mod;
   }
    for(int i=2;i<=k;++i)
    {
        for(int j=2;j<=n;++j)
        {   
            
            
            dp[i][j]=cum[i][j-1]%mod;
            if(min(i,a[j-1])!=i and i-min(i,a[j-1])!=0)
            {
               // cout<<i<<" "<<j<<" "<<min(i,a[j-1])<<endl;
                dp[i][j]=(dp[i][j]-cum[i-min(i,a[j-1])-1][j-1]+mod)%mod;
               
            }
            dp[i][j]%=mod;
            cum[i][j]=(cum[i-1][j]+dp[i][j]+mod)%mod;
            
            //cout<<dp[i][j]<<" "<<i<<" "<<j<<endl;
            // if(i==2 and j==1)
            //     cout<<dp[i][j]<<endl;
            
               //  for(int w=0;w<=min(i,a[j-1]);++w)
               //      dp[i][j]+=dp[i-w][j-1];
               //  dp[i][j]%=mod;
               //  if(j==1)
               // cout<<i<<" "<<j<<" "<<dp[i][j]<<endl;
            
            
        }

    }
    cout<<dp[k][n]%mod;
}