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)

import java.io.{BufferedReader, InputStream, InputStreamReader, PrintWriter}
import java.util.StringTokenizer

import scala.collection.mutable
import scala.util.Sorting
import math.{abs, max, min}
import scala.collection.mutable.{ArrayBuffer, ListBuffer}
import scala.reflect.ClassTag

object Main {
  val MOD = 1000000007
  val out = new PrintWriter(System.out)
  def solve(): Unit = {
    val sc = new InputReader(System.in)
    val N, C = sc.nextInt()
    val A, B = map(N)(_ => sc.nextInt())
    val X = Array.ofDim[Long](401, C + 1)
    for {
      i <- 0 to 400
      j <- 0 to C
    } X(i)(j) = pow_mod(i, j, MOD)

    for {
      i <- 1 to 400
      j <- 0 to C
    } {
      X(i)(j) += X(i - 1)(j)
      X(i)(j) %= MOD
    }

    val dp = Array.ofDim[Long](N + 1, C + 1)
    dp(0)(0) = 1
    for {
      i <- 0 until N
      j <- 0 to C
      k <- 0 to j
    } {
      dp(i + 1)(j) += dp(i)(j - k) * (X(B(i))(k) - X(A(i) - 1)(k) + MOD)
      dp(i + 1)(j) %= MOD
    }

    val ans = dp(N)(C)
    out.println(ans)
  }


  def main(args: Array[String]): Unit = {
    solve()
    out.flush()
  }


  class InputReader(val stream: InputStream) {
    private val reader = new BufferedReader(new InputStreamReader(stream), 32768)
    private var tokenizer: StringTokenizer = null

    def next: String = {
      while (tokenizer == null || !tokenizer.hasMoreTokens)
        tokenizer = new StringTokenizer(reader.readLine)
      tokenizer.nextToken
    }

    def nextInt(): Int = next.toInt
    def nextLong(): Long = next.toLong
    def nextChar(): Char = next.charAt(0)
  }

  private def rep(n: Int)(f: Int => Unit): Unit = (0 until n) foreach f

  private def map[@specialized A: ClassTag](n: Int)(f: Int => A): Array[A] = {
    val res = Array.ofDim[A](n)
    (0 until n) foreach (i => res(i) = f(i))
    res
  }

  private implicit class ArrayOpts[A](val as: Array[A]) extends AnyVal {
    def maxByOpt[B: Ordering](f: A => B): Option[A] = {
      if (as.nonEmpty) Some(as.maxBy(f)) else None
    }

    def grpBy[K](f: A => K): mutable.Map[K, ArrayBuffer[A]] = {
      val map = mutable.Map.empty[K, ArrayBuffer[A]]
      rep(a => map.getOrElseUpdate(f(a), ArrayBuffer()) += a)
      map
    }

    def rep(f: A => Unit): Unit = {
      as.indices foreach (i => f(as(i)))
    }

    def sumBy[B](f: A => B)(implicit num: Numeric[B]): B = {
      var sum = num.zero
      rep(a => sum = num.plus(sum, f(a)))
      sum
    }
  }

  private implicit class IterableOpts[A](val as: Iterable[A]) extends AnyVal {
    def sumBy[B](f: A => B)(implicit num: Numeric[B]): B = {
      as.foldLeft(num.zero)((acc, a) => num.plus(acc, f(a)))
    }
  }

  def pow_mod(x: Int, n: Int, m: Int): Int = {
    def step(x0: Long, n0: Int): Long = {
      n0 match {
        case 0 => 1
        case _ =>
          val r = step(x0 * x0 % m, n0 / 2)
          if (n0 % 2 == 1) r * x0 % m else r
      }
    }
    step(x.toLong, n).toInt
  }
}