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)

#![allow(unused_imports, unused_variables, dead_code)]
use std::io::*;
use std::fmt::*;
use std::str::*;
use std::cmp::*;
use std::collections::*;

trait InputValue {
    fn parse(s: &str) -> Self;
}

fn read<T: InputValue>() -> T {
    let mut buf = String::new();
    let _ = stdin().read_line(&mut buf);
    T::parse(&buf.trim())
}

fn readnc<T: InputValue>() -> Vec<T> {
    let mut vec = vec![];
    let line: String = read();
    for token in line.split_whitespace() {
        vec.push(T::parse(token));
    }
    vec
}

fn readn<T: InputValue>(n: usize) -> Vec<T> {
    let mut vec = vec![];
    for _ in 0..n {
        vec.push(read());
    }
    vec
}

macro_rules! parse_single_value {
    ($($t:ty),*) => {
        $(
            impl InputValue for $t {
                fn parse(s: &str) -> $t { s.parse().unwrap() }
            }
        )*
	}
}
parse_single_value!(i32, i64, f32, f64, usize, String);

macro_rules! parse_tuple {
	($($t:ident),*) => {
		impl<$($t),*> InputValue for ($($t),*) where $($t: InputValue),* {
			fn parse(s: &str) -> ($($t),*) {
				let mut tokens = s.split_whitespace();
				let t = ($($t::parse(tokens.next().unwrap())),*);
				t
			}
		}
	}
}
parse_tuple!(A, B);
parse_tuple!(A, B, C);

// ===

const MOD: i64 = 1000000007;

fn powmod(a: i64, p: i64, m: i64) -> i64 {
    let mut ret = 1i64;
    let mut aa = a;
    let mut pp = p;
    while pp >= 1 {
        if pp & 1 == 1 {
            ret *= aa;
            ret %= m;
        }
        aa = aa * aa % m;
        pp >>= 1;
    }
    ret
}


fn main() {
    let (n, c): (usize, usize) = read();
    let a: Vec<usize> = readnc();
    let b: Vec<usize> = readnc();

    let mut ranges: Vec<(usize, usize)> = vec![];
    for i in 0..n {
        ranges.push((a[i], b[i]));
    }

    let mut pow_k: Vec<Vec<i64>> = vec![vec![0; 401]; 401];
    for k in 0..401 {
        let mut sum: i64 = 0;
        for i in 1..401 {
            sum += powmod(i, k, MOD);
            if sum >= MOD {
                sum -= MOD;
            }
            pow_k[k as usize][i as usize] = sum;
        }
    }

    let mut memo: Vec<Vec<i64>> = vec![vec![-1; c+1]; n+1];

    println!("{}", dfs(0, c, &ranges, &pow_k, &mut memo));
}

fn dfs(i: usize, left: usize, range: &Vec<(usize, usize)>, pow_k: &Vec<Vec<i64>>, memo: &mut Vec<Vec<i64>>) -> i64 {
    if memo[i][left] != -1 {
        return memo[i][left]
    }
    if i == range.len() {
        let ret = if left == 0 { 1 } else { 0 };
        memo[i][left] = ret;
        return ret;
    }

    let rng = range[i];
    let mut sum = 0;
    for u in 0..left+1 {
        let mut p = pow_k[u][rng.1] - pow_k[u][rng.0-1] + MOD;
        if p >= MOD {
            p -= MOD;
        }
        sum += p * dfs(i+1, left-u, range, pow_k, memo) % MOD;
        if sum >= MOD {
            sum -= MOD;
        }
    }
    memo[i][left] = sum;
    sum
}