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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;
import java.io.*;

 public class Main{
static BufferedReader s1;
static BufferedWriter out;
static String read() throws IOException{String line="";while(line.length()==0){line=s1.readLine();continue;}return line;}
static int int_v (String s1){return Integer.parseInt(s1);}
static long long_v(String s1){return Long.parseLong(s1);}
static int[] int_arr() throws IOException{String[] a=read().split(" ");int[] b=new int[a.length];for(int i=0;i<a.length;i++){b[i]=int_v(a[i]);}return b;}
static long[] long_arr() throws IOException{String[] a=read().split(" ");long[] b=new long[a.length];for(int i=0;i<a.length;i++){b[i]=long_v(a[i]);}return b;}
static void assign(){s1=new BufferedReader(new InputStreamReader(System.in));out=new BufferedWriter(new OutputStreamWriter(System.out));}
//......................................@uthor_Alx..............................................

   public static void main(String[] args) throws  IOException{
	        		  assign();
	        		  int[] x=int_arr();
	        		  int n=x[0],k=x[1];
	        		  long[][] prefix=new long[n+1][k+1];
	        		  long mod=(long)1e9+7;
	        		  int[] a=int_arr();
	 
	        		  /*
	        		  # prefix[i][j] holds sum of coeff of x^0,x^1,....,x^j  in product
	        		  of 1st i polynomials

	        		  # coeff of X^k in product of 1st i polynomials therefore depends on
	        		  whether highest degree of X (a[i] here) in last polynomial i.e. "ith" poly is
	        		  >= k or not.

	        		  Case1:- if a[i]>=k then all it's term till (x^0,x^1,....,x^k) will contribute
	        		  (see math) so coeff of x^k =prefix[i-1][k]

	        		  Case2:- a[i]<k then all it's term (x^0,x^1,....,x^a[i]) will contribute so 
	        		  in previous product only term contributing are (x^k,x^(k-1),....,x^(k-a[i]))
	        		   so coeff of x^k =prefix[i-1][k]-prefix[i-1][k-a[i]-1];
	        		  */
                      Arrays.fill(prefix[0],1);
	        		  for(int i=1;i<=n;i++){prefix[i][0]=1;}
	        		  for(int i=1;i<=n;i++){
	        		  	  for(int j=1;j<=k;j++){
	        		  	  	long r=(j-a[i-1]-1>=0)?prefix[i-1][j-a[i-1]-1]:0;
	        		  	  	long l=prefix[i-1][j]; 
	        		  	  	prefix[i][j]=(prefix[i][j-1]+(l-r+mod)%mod)%mod;
	        		  	  }
	        		  }
	        		  long r=(k-a[n-1]-1>=0)?prefix[n-1][k-a[n-1]-1]:0;
	        		  long l=prefix[n-1][k]; 
	        		  out.write((l-r+mod)%mod+"");
                      out.flush();
	        		 
	}
}