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 Pascal (FPC 2.6.2) (judge verdict: Wrong Answer)

var a: array[1..1000] of longint;
    dp, p: array[-10..150, -110000..200000] of longint;    
    i, j, N, K: longint;
begin
    readln(N, K);
    for i:=1 to N do
        read(a[i]);
    for i:=0 to a[1] do
    begin
        dp[1,i]:=1;
    end;
    for i:=0 to K do
        p[1,i]:=p[1,i-1]+dp[1,i];
    for i:=2 to N do
    begin
        for j:=0 to K do
        begin
            dp[i,j]:=(p[i-1,j]-p[i-1,j-a[i]-1]) mod 1000000007;
            p[i,j]:=(p[i,j-1]+dp[i,j]) mod 1000000007;
//            write(dp[i,j], ' ');
        end;
//        writeln;
    end;
    writeln(dp[N,K]);
end.