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.

Examples

Input

3 4
1 2 3


Output

5


Input

1 10
9


Output

0


Input

2 0
0 0


Output

1


Input

4 100000
100000 100000 100000 100000


Output

665683269
import bisect
import copy
import heapq
import math
import sys
from collections import *
from itertools import accumulate, combinations, permutations, product
# from math import gcd
from functools import lru_cache
def input():
    return sys.stdin.readline()[:-1]
def ruiseki(lst):
    return [0]+list(accumulate(lst))
sys.setrecursionlimit(5000000)
mod=pow(10,9)+7
al=[chr(ord('a') + i) for i in range(26)]
direction=[[1,0],[0,1],[-1,0],[0,-1]]

n,k=map(int,input().split())
a=list(map(int,input().split()))

dp=[[0]*(k+1) for j in range(n+1)]

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

for i in range(1,n+1):
    for j in range(1,k+1):
        dp[i][j]=dp[i-1][j]+dp[i][j-1]
        if j-a[i-1]-1 >= 0:
            dp[i][j]-=dp[i-1][j-a[i-1]-1]
        dp[i][j]%=mod
# print(dp)
print(dp[n][k])