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 Rust (1.15.1) (judge verdict: Accepted)

use std::io::BufRead;

const P: u64 = 1_000_000_007;

fn main() {
    let stdin = std::io::stdin();
    let (n, c): (usize, usize) = stdin.lock().gets().split(' ').values();
    let a = stdin.lock().gets().split(' ').map(|x| x.parse().unwrap())
        .collect::<Vec<u64>>();
    let b = stdin.lock().gets().split(' ').map(|x| x.parse().unwrap())
        .collect::<Vec<u64>>();

    let mut cnt = vec![vec![0; c + 1]; n + 1];
    cnt[0][0] = 1;
    for ni in 1..n + 1 {
        for x in a[ni - 1]..b[ni - 1] + 1 {
            let mut cnt_ni = vec![0; c + 1];
            cnt_ni[0] = cnt[ni - 1][0];
            for ci in 1..c + 1 {
                cnt_ni[ci] = (x * cnt_ni[ci - 1] + cnt[ni - 1][ci]) % P;
            }
            for ci in 0..c + 1 {
                cnt[ni][ci] += cnt_ni[ci];
            }
        }
        for ci in 0..c + 1 {
            cnt[ni][ci] %= P;
        }
    }
    println!("{}", cnt[n][c]);
}

trait Gets { fn gets(&mut self) -> String; }
impl<T: std::io::BufRead> Gets for T {
    fn gets(&mut self) -> String { self.lines().next().unwrap().unwrap() } }

trait Values<T> { fn values(&mut self) -> T; }
impl<'a, I, T1, T2> Values<(T1, T2)> for I
    where I: Iterator<Item=&'a str>,
          T1: std::str::FromStr, T1::Err: std::fmt::Debug,
          T2: std::str::FromStr, T2::Err: std::fmt::Debug {
    fn values(&mut self) -> (T1, T2) {
        (self.next().unwrap().parse().unwrap(),
         self.next().unwrap().parse().unwrap())
    }
}