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))
    }
}

impl From<usize> for ModInt {
    fn from(val: usize) -> ModInt {
        ModInt((val % MOD as usize) as u32)
    }
}

#[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 {
        assert!(self.0 > 0);
        self.pow(MOD - 2)
    }
}
// ---------- end ModInt ----------
// ---------- begin Precalc ----------
#[allow(dead_code)]
struct Precalc {
    inv: Vec<ModInt>,
    fact: Vec<ModInt>,
    ifact: Vec<ModInt>,
}

#[allow(dead_code)]
impl Precalc {
    pub fn new(n: usize) -> Precalc {
        let mut inv = vec![ModInt::one(); n + 1];
        let mut fact = vec![ModInt::one(); n + 1];
        let mut ifact = vec![ModInt::one(); n + 1];
        for i in 2..(n + 1) {
            inv[i] = -inv[MOD as usize % i] * ModInt(MOD / i as u32);
            fact[i] = fact[i - 1] * ModInt(i as u32);
            ifact[i] = ifact[i - 1] * inv[i];
        }
        Precalc {
            inv: inv,
            fact: fact,
            ifact: ifact,
        }
    }
    pub fn inv(&self, n: usize) -> ModInt {
        self.inv[n]
    }
    pub fn fact(&self, n: usize) -> ModInt {
        self.fact[n]
    }
    pub fn ifact(&self, n: usize) -> ModInt {
        self.ifact[n]
    }
    pub fn comb(&self, n: usize, k: usize) -> ModInt {
        if k > n {
            return ModInt::zero();
        }
        self.fact[n] * self.ifact[k] * self.ifact[n - k]
    }
}
// ---------- end Precalc ----------
//https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8 より
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
    ($($r:tt)*) => {
        let s = {
            use std::io::Read;
            let mut s = String::new();
            std::io::stdin().read_to_string(&mut s).unwrap();
            s
        };
        let mut iter = s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
}

macro_rules! input_inner {
    ($iter:expr) => {};
    ($iter:expr, ) => {};
    ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };
}

macro_rules! read_value {
    ($iter:expr, ( $($t:tt),* )) => {
        ( $(read_value!($iter, $t)),* )
    };
    ($iter:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
    };
    ($iter:expr, chars) => {
        read_value!($iter, String).chars().collect::<Vec<char>>()
    };
    ($iter:expr, usize1) => {
        read_value!($iter, usize) - 1
    };
    ($iter:expr, $t:ty) => {
        $iter.next().unwrap().parse::<$t>().expect("Parse error")
    };
}

//

fn run() {
    input! {
        n: usize,
        c: usize,
        a: [usize; n],
        b: [usize; n],
    }
    let m = 400;
    let mut sum = vec![vec![ModInt::zero(); m + 1]; m + 1];
    for (i, sum) in sum.iter_mut().enumerate() {
        for (j, s) in sum.iter_mut().enumerate() {
            *s = ModInt::from(i).pow(j as u32);
        }
    }
    for i in 1..(m + 1) {
        let (x, y) = sum.split_at_mut(i);
        for (x, y) in x.last().unwrap().iter().zip(y[0].iter_mut()) {
            *y += *x;
        }
    }
    let mut dp = vec![ModInt::zero(); c + 1];
    dp[0] = ModInt::one();
    for (a, b) in a.into_iter().zip(b.into_iter()) {
        let mut next = vec![ModInt::zero(); c + 1];
        for (i, v) in dp.into_iter().enumerate() {
            for (k, (b, a)) in sum[b].iter().zip(sum[a - 1].iter()).enumerate() {
                if i + k > c {
                    break;
                }
                next[i + k] += v * (*b - *a);
            }
        }
        dp = next;
    }
    let ans = dp[c];
    println!("{}", ans);
}

fn main() {
    run();
}