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)

const MOD: usize = 1e9 as usize + 7;
fn main() {
    let s = std::io::stdin();
    let mut sc = Scanner { stdin: s.lock() };

    let n = sc.read();
    let c: usize = sc.read();
    let a: Vec<usize> = sc.vec(n);
    let b: Vec<usize> = sc.vec(n);
    let max_b: usize = *b.iter().max().unwrap();

    let mut pow = vec![vec![0; c + 1]; max_b + 1];
    for x in 0..(max_b + 1) {
        pow[x][0] = 1;
        for i in 0..c {
            pow[x][i + 1] = (pow[x][i] * x) % MOD;
        }
    }

    let mut sum_pow = vec![vec![0; c + 1]; max_b + 2];
    for i in 0..(c + 1) {
        for x in 0..(max_b + 1) {
            sum_pow[x + 1][i] = (sum_pow[x][i] + pow[x][i]) % MOD;
        }
    }

    let mut dp = vec![0; c + 1];
    dp[0] = 1;
    for i in 0..n {
        let (a, b) = (a[i], b[i]);
        let mut next = vec![0; c + 1];
        for cur in 0..(c + 1) {
            if dp[cur] == 0 {
                continue;
            }
            for add in 0..(c + 1) {
                if cur + add > c {
                    break;
                }

                next[cur + add] += dp[cur] * (sum_pow[b + 1][add] + MOD - sum_pow[a][add]);
                next[cur + add] %= MOD;
            }
        }
        dp = next;
    }
    println!("{}", dp[c]);
}

pub struct Scanner<R> {
    stdin: R,
}

impl<R: std::io::Read> Scanner<R> {
    pub fn read<T: std::str::FromStr>(&mut self) -> T {
        use std::io::Read;
        let buf = self
            .stdin
            .by_ref()
            .bytes()
            .map(|b| b.unwrap())
            .skip_while(|&b| b == b' ' || b == b'\n' || b == b'\r')
            .take_while(|&b| b != b' ' && b != b'\n' && b != b'\r')
            .collect::<Vec<_>>();
        unsafe { std::str::from_utf8_unchecked(&buf) }
            .parse()
            .ok()
            .expect("Parse error.")
    }
    pub fn vec<T: std::str::FromStr>(&mut self, n: usize) -> Vec<T> {
        (0..n).map(|_| self.read()).collect()
    }
    pub fn chars(&mut self) -> Vec<char> {
        self.read::<String>().chars().collect()
    }
}