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: Time Limit Exceeded)

#include <iostream>
#define mod 1000000007
#define MAX 100001
using namespace std;


int findit(int currentindex,int arr[],int n,int k,int dp[][MAX])
{
    
    if(k<=0)
    {
        return 1;
    }
    
    if(dp[currentindex][k]!=-1)
    {
        return dp[currentindex][k];
    }
    
    if(currentindex==n-1)
    {
        if(k<=arr[currentindex])
        {
            dp[currentindex][k]=1;
            return dp[currentindex][k];
        }
        return 0;
    }
    
    int ans=0;
    
    for(int i=0;i<=arr[currentindex];i++)
    {
        int nextcount=0;
        
        if(k-i<=0)  //0 1 i=2 3
        {
            nextcount=arr[currentindex]-i+1;
            ans=( (ans % mod) + (nextcount%mod) ) %mod;
            break;
        }
        
       
        nextcount=findit(currentindex+1,arr,n,k-i,dp)%mod;
        ans=( (ans % mod) + (nextcount%mod) ) %mod;
        
    }
    
    dp[currentindex][k]=ans;
    return dp[currentindex][k];
    
}


int main() 
{
int n;
int k;

cin>>n;
cin>>k;

int arr[n];
int dp[n][MAX];
for(int i=0;i<n;i++)
{
    for(int j=0;j<MAX;j++)
    {
        dp[i][j]=-1;
    }
}

for(int i=0;i<n;i++)
{
    cin>>arr[i];
}

cout<<findit(0,arr,n,k,dp);

	return 0;
}