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 Ruby (2.3.3) (judge verdict: Time Limit Exceeded)

N,K=gets.chomp.split.map(&:to_i)
a=gets.chomp.split.map(&:to_i)

dp=Array.new(N+1).map{Array.new(K+1,0)}

MOD=1000000007

dp[0][0] = 1

csum = Array.new(K+1)
1.upto(N) do |i|
  csum[0] = dp[i-1][0]
  1.upto(K) {|j| csum[j] = csum[j-1]+dp[i-1][j]}
  0.upto(K) do |j|
    if j-a[i-1]-1 >= 0
      dp[i][j] = csum[j] - csum[j-a[i-1]-1]   
    else
      dp[i][j] = csum[j]
    end
    dp[i][j] %= MOD
  end
end

puts dp[N][K]