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)

// ---------- begin ModInt ----------
const MOD: u32 = 1_000_000_007;

#[derive(Clone, Copy)]
struct ModInt(u32);

impl std::ops::Add for ModInt {
    type Output = ModInt;
    fn add(self, rhs: ModInt) -> Self::Output {
        let mut d = self.0 + rhs.0;
        if d >= MOD {
            d -= MOD;
        }
        ModInt(d)
    }
}

impl std::ops::AddAssign for ModInt {
    fn add_assign(&mut self, rhs: ModInt) {
        *self = *self + rhs;
    }
}

impl std::ops::Sub for ModInt {
    type Output = ModInt;
    fn sub(self, rhs: ModInt) -> Self::Output {
        let mut d = self.0 + MOD - rhs.0;
        if d >= MOD {
            d -= MOD;
        }
        ModInt(d)
    }
}

impl std::ops::SubAssign for ModInt {
    fn sub_assign(&mut self, rhs: ModInt) {
        *self = *self - rhs;
    }
}

impl std::ops::Mul for ModInt {
    type Output = ModInt;
    fn mul(self, rhs: ModInt) -> Self::Output {
        ModInt((self.0 as u64 * rhs.0 as u64 % MOD as u64) as u32)
    }
}

impl std::ops::MulAssign for ModInt {
    fn mul_assign(&mut self, rhs: ModInt) {
        *self = *self * rhs;
    }
}

impl std::ops::Neg for ModInt {
    type Output = ModInt;
    fn neg(self) -> Self::Output {
        ModInt(if self.0 == 0 {0} else {MOD - self.0})
    }
}

impl std::fmt::Display for ModInt {
    fn fmt<'a>(&self, f: &mut std::fmt::Formatter<'a>) -> std::fmt::Result {
        write!(f, "{}", self.0)
    }
}

impl std::str::FromStr for ModInt {
    type Err = std::num::ParseIntError;
    fn from_str(s: &str) -> Result<Self, Self::Err> {
        let val = s.parse::<u32>()?;
        Ok(ModInt::new(val))
    }
}

#[allow(dead_code)]
impl ModInt {
    pub fn new(n: u32) -> ModInt {
        ModInt(n % MOD)
    }
    pub fn zero() -> ModInt {
        ModInt(0)
    }
    pub fn one() -> ModInt {
        ModInt(1)
    }
    pub fn pow(self, mut n: u32) -> ModInt {
        let mut t = ModInt::one();
        let mut s = self;
        while n > 0 {
            if n & 1 == 1 {
                t *= s;
            }
            s *= s;
            n >>= 1;
        }
        t
    }
    pub fn inv(self) -> ModInt {
        self.pow(MOD - 2)
    }
}
// ---------- end ModInt ----------

use std::io::Read;

fn run() {
    let mut s = String::new();
    std::io::stdin().read_to_string(&mut s).unwrap();
    let mut it = s.trim().split_whitespace();
    let n: usize = it.next().unwrap().parse().unwrap();
    let c: usize = it.next().unwrap().parse().unwrap();
    let a: Vec<usize> = it.map(|s| s.parse().unwrap()).collect();
    let (a, b) = a.split_at(n);
    let m = 400;
    let mut sum = vec![vec![ModInt::zero(); c + 1]; m + 1];
    for (i, s) in sum.iter_mut().enumerate() {
        for (j, s) in s.iter_mut().enumerate() {
            *s = ModInt(i as u32).pow(j as u32);
        }
    }
    for i in 1..(m + 1) {
        for j in 0..(c + 1) {
            sum[i][j] = sum[i][j] + sum[i - 1][j];
        }
    }
    let mut dp = vec![ModInt::zero(); c + 1];
    dp[0] = ModInt::one();
    for (&a, &b) in a.iter().zip(b.iter()) {
        let mut next = vec![ModInt::zero(); c + 1];
        for (i, dp) in dp.iter().enumerate() {
            for (next, (x, y)) in next[i..].iter_mut().zip(sum[b].iter().zip(sum[a - 1].iter())) {
                *next += *dp * (*x - *y);
            }
        }
        dp = next;
    }
    println!("{}", dp[c]);
}

fn main() {
    run();
}