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 Scala (2.11.7) (judge verdict: Wrong Answer)

import java.io._
import java.util.StringTokenizer
import scala.util.control.Breaks.{breakable,break}
import scala.collection.mutable.{Map, Queue, PriorityQueue}

object Main extends App {
  val in = new InputReader(System.in)
  val pw = new PrintWriter(System.out)

  val N, K = in.next().toInt
  val a = new Array[Int](N+1)
  for (i <- 1 to N)
    a(i) = in.next().toInt


  //
  val MOD = 1e9.toInt + 7
  val dp = Array.ofDim[Long](N+2, K+2)
  dp(0)(0) = 1

  for (i <- 1 to N) {
    for (j <- 1 to K) {
      dp(i-1)(j) += dp(i-1)(j-1)
      dp(i-1)(j) %= MOD
    }

    val now = a(i)
    for (k <- 0 to now) {
      dp(i)(k) = dp(i-1)(k)
      dp(i)(k) %= MOD
    }
    for (k <- now+1 to K) {
      dp(i)(k) = dp(i-1)(k) - dp(i-1)(k-(now+1))
      dp(i)(k) %= MOD
    }
  }

  println(dp(N)(K))
}


class InputReader(stream: InputStream) {
  private val reader = new BufferedReader(new InputStreamReader(stream))
  private var tokenizer: StringTokenizer = new StringTokenizer(reader.readLine())
  def next(): String = {
    while (!tokenizer.hasMoreTokens()) {
      tokenizer = new StringTokenizer(reader.readLine())
    }
    tokenizer.nextToken()
  }
}