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 PyPy2 (5.6.0) (judge verdict: Time Limit Exceeded)

# -*- coding: utf-8 -*-
import sys

N,K=map(int, sys.stdin.readline().split())
A=map(int, sys.stdin.readline().split())

dp=[ [] for i in range(N+1) ]
dp[0]=[1]

mod=10**9+7


for n in range(N):
	W=[0]		#WA:累積和
	L=dp[n]
	l=len(L)
	a=A[n]
	for j,x in enumerate(L):
		if j==0:	W.append(x)
		else:	W.append( W[-1]+x )

	T=[]  #Temporary
	for k in range( min (K+1,a+l) ):
		#print k, min( k+1, l), max(0,k-a), W[min( k+1, l)]-W[max(0,k-a)]
		T.append( W[min( k+1, l)]-W[max(0,k-a)] )

	dp[n+1]=T



try:
	print dp[N][K]%mod
except IndexError:
	print 0