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 Go (1.6) (judge verdict: Wrong Answer)

package main

import (
  "fmt"
  "math"
)

func max(M *[]int) (ret int) {
  ret = (*M)[0]
  for _, m := range (*M) {
    if ret < m {
      ret = m
    }
  }
  return
}

func main() {
  var N, C int
  fmt.Scan(&N, &C)
  A := make([]int, N)
  B := make([]int, N)
  p := int64(math.Pow(10, 9) + 7)
  var x int
  for i := 0; i < N; i++ {
    fmt.Scan(&x)
    A[i] = x
  }
  for i := 0; i < N; i++ {
    fmt.Scan(&x)
    B[i] = x
  }
  B_max := max(&B)
  X := make([][]int64, C + 2)
  for i := 0; i < C + 2; i++ {
    X[i] = make([]int64, B_max + 1)
  }
  for j := 1; j < B_max + 1; j++ {
    X[0][j] = 1
  }
  for i := 0; i < C + 1; i++ {
    for j := 1; j < B_max + 1; j++ {
      X[i + 1][j] = (X[i][j] * int64(j)) % p
      X[i][j] = (X[i][j] + X[i][j - 1]) % p
    }
  }
  DP := make([][]int64, N + 1)
  for i := 0; i < N + 1; i++ {
    DP[i] = make([]int64, C + 1)
  }
  DP[0][0] = 1
  for i := 1; i < N + 1; i++ {
    for j := 0; j < C + 1; j++ {
      for k := 0; k < j + 1; k++ {
        DP[i][j] += (DP[i - 1][j - k] * (X[k][B[i - 1]] - X[k][A[i - 1] - 1])) % p
        DP[i][j] %= p
      }
    }
  }
  fmt.Println(DP[N][C])
}