Problem:
Score : 800 points

Problem Statement

12:17 (UTC): The sample input 1 and 2 were swapped. The error is now fixed. We are very sorry for your inconvenience.

There are N children in AtCoder Kindergarten, conveniently numbered 1 through N. Mr. Evi will distribute C indistinguishable candies to the children.

If child i is given a candies, the child's happiness will become x_i^a, where x_i is the child's excitement level. The activity level of the kindergarten is the product of the happiness of all the N children.

For each possible way to distribute C candies to the children by giving zero or more candies to each child, calculate the activity level of the kindergarten. Then, calculate the sum over all possible way to distribute C candies. This sum can be seen as a function of the children's excitement levels x_1,..,x_N, thus we call it f(x_1,..,x_N).

You are given integers A_i,B_i (1≦i≦N). Find modulo 10^9+7.

Constraints

1≦N≦400

1≦C≦400

1≦A_i≦B_i≦400 (1≦i≦N)

Partial Score

400 points will be awarded for passing the test set satisfying A_i=B_i (1≦i≦N).

Input

The input is given from Standard Input in the following format:

N C
A_1 A_2 ... A_N
B_1 B_2 ... B_N

Output

Print the value of modulo 10^9+7.

Sample Input 1

2 3
1 1
1 1

Sample Output 1

4

This case is included in the test set for the partial score, since A_i=B_i.
We only have to consider the sum of the activity level of the kindergarten where the excitement level of both child 1 and child 2 are 1 (f(1,1)).

If child 1 is given 0 candy, and child 2 is given 3 candies, the activity level of the kindergarten is 1^0*1^3=1.

If child 1 is given 1 candy, and child 2 is given 2 candies, the activity level of the kindergarten is 1^1*1^2=1.

If child 1 is given 2 candies, and child 2 is given 1 candy, the activity level of the kindergarten is 1^2*1^1=1.

If child 1 is given 3 candies, and child 2 is given 0 candy, the activity level of the kindergarten is 1^3*1^0=1.

Thus, f(1,1)=1+1+1+1=4, and the sum over all f is also 4.

Sample Input 2

1 2
1
3

Sample Output 2

14

Since there is only one child, child 1's happiness itself will be the activity level of the kindergarten. Since the only possible way to distribute 2 candies is to give both candies to child 1, the activity level in this case will become the value of f.

When the excitement level of child 1 is 1, f(1)=1^2=1.

When the excitement level of child 1 is 2, f(2)=2^2=4.

When the excitement level of child 1 is 3, f(3)=3^2=9.

Thus, the answer is 1+4+9=14.

Sample Input 3

2 3
1 1
2 2

Sample Output 3

66

Since it can be seen that f(1,1)=4 , f(1,2)=15 , f(2,1)=15 , f(2,2)=32, the answer is 4+15+15+32=66.

Sample Input 4

4 8
3 1 4 1
3 1 4 1

Sample Output 4

421749

This case is included in the test set for the partial score.

Sample Input 5

3 100
7 6 5
9 9 9

Sample Output 5

139123417

Solution in Scala (2.11.7) (judge verdict: Accepted)

object Main extends App {
  import scala.io.StdIn._

  val mod = 1000000007L

  val Array(n, c) = readLine.trim.split(' ').map(_.toInt)
  val a = readLine.trim.split(' ').map(_.toInt)
  val b = readLine.trim.split(' ').map(_.toInt)

  println(
    solve(n, c, a, b)
  )

  def solve(n: Int, c: Int, a: Array[Int], b: Array[Int]): Long = {
    val memo = Array.tabulate(n){_ ⇒ Array.tabulate(c + 1){_ ⇒ -1L}}
    val sumExp = Array.tabulate(401){_ ⇒ Array.tabulate(401){_ ⇒ 0L}}.also{arr ⇒
      for (ab ← arr.indices){
        arr(ab)(0) = 1
        for (ar ← arr(ab).indices.tail){
          arr(ab)(ar) = (arr(ab)(ar - 1) * ab) % mod
        }
      }
      for (i ← arr(0).indices) arr(0)(i) = 0
      for (ab ← arr.indices.tail){
        for (ar ← arr(ab).indices){
          arr(ab)(ar) = (arr(ab)(ar) + arr(ab - 1)(ar)) % mod
        }
      }
    }
    def sub(position: Int, rest: Int): Long = {
      if (position == n) {
        if (rest == 0) 1L
        else 0L
      }else {
        if (memo(position)(rest) == -1L){
          memo(position)(rest) = 0
          for (use ← 0 to rest){
            memo(position)(rest) = (memo(position)(rest) + ((sumExp(b(position))(use) - sumExp(a(position) - 1)(use) + mod) % mod) * sub(position + 1, rest - use)) % mod
          }
        }
        memo(position)(rest)
      }
    }
    sub(0, c)
  }
  def exponent(base: Long, exp: Int): Long = {
    if (exp == 0) 1
    else if (exp == 1) base
    else exponent(base, exp / 2).let(it ⇒ it * it * exponent(base, exp % 2)) % mod
  }


  implicit class Scope[T](val value: T) extends AnyVal {
    @inline
    def let[A](func: T ⇒ A): A = func(value)
    @inline
    def also(func: T ⇒ Unit): T = {
      func(value)
      value
    }
  }
}