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 Go (1.6) (judge verdict: Accepted)

package main

import (
    "fmt"
)

var n, k int 
var as []int
var dp [][]int
var sdp [][]int

const mod = int(1e9 + 7)

func main() {
    fmt.Scan(&n, &k) 
    as = make([]int, n)
    for i := range as {
        var a int 
        fmt.Scan(&a)
        as[i] = a 
    }   

    // dp[i][k] i番目まででk個になる通り
    dp = make([][]int, n)
    sdp = make([][]int, n)
    for i := 0; i < n; i++ {
        dp[i] = make([]int, k+1)
        sdp[i] = make([]int, k+2)
    }   

    dp[0][0] = 1 
    sdp[0][1] = 1 
    for j := 1; j < k+1; j++ {
        if j <= as[0] {
            dp[0][j] = 1 
        }   
        sdp[0][j+1] = sdp[0][j] + dp[0][j]
    }   

    for i := 1; i < n; i++ {
        for j := 0; j < k+1; j++ {
            dp[i][j] = gm(sdp[i-1][j+1] - sdp[i-1][max(j-as[i], 0)])
            sdp[i][j+1] = gm(sdp[i][j] + dp[i][j])
        }   
    }   

    fmt.Println(dp[n-1][k])
}

func max(a, b int) int {
    if a > b { 
        return a
    }   
    return b
}

func gm(n int) int {
    if n < 0 { 
        n += mod 
    }   
    return n % mod 
}