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: Wrong Answer)

#![allow(unused_imports)]
#![allow(non_snake_case, unused)]

use std::cmp::*;
use std::collections::*;
use std::ops::*;

// https://atcoder.jp/contests/hokudai-hitachi2019-1/submissions/10518254

macro_rules! eprint {
	($($t:tt)*) => {{
		use ::std::io::Write;
		let _ = write!(::std::io::stderr(), $($t)*);
	}};
}
macro_rules! eprintln {
	() => { eprintln!(""); };
	($($t:tt)*) => {{
		use ::std::io::Write;
		let _ = writeln!(::std::io::stderr(), $($t)*);
	}};
}
macro_rules! dbg {
	($v:expr) => {{
		let val = $v;
		eprintln!("[{}:{}] {} = {:?}", file!(), line!(), stringify!($v), val);
		val
	}}
}

macro_rules! mat {
	($($e:expr),*) => { Vec::from(vec![$($e),*]) };
	($($e:expr,)*) => { Vec::from(vec![$($e),*]) };
	($e:expr; $d:expr) => { Vec::from(vec![$e; $d]) };
	($e:expr; $d:expr $(; $ds:expr)+) => { Vec::from(vec![mat![$e $(; $ds)*]; $d]) };
}

macro_rules! ok {
	($a:ident$([$i:expr])*.$f:ident()$(@$t:ident)*) => {
		$a$([$i])*.$f($($t),*)
	};
	($a:ident$([$i:expr])*.$f:ident($e:expr$(,$es:expr)*)$(@$t:ident)*) => { {
		let t = $e;
		ok!($a$([$i])*.$f($($es),*)$(@$t)*@t)
	} };
}

pub fn readln() -> String {
	let mut line = String::new();
	::std::io::stdin().read_line(&mut line).unwrap_or_else(|e| panic!("{}", e));
	line
}

macro_rules! read {
	($($t:tt),*; $n:expr) => {{
		let stdin = ::std::io::stdin();
		let ret = ::std::io::BufRead::lines(stdin.lock()).take($n).map(|line| {
			let line = line.unwrap();
			let mut it = line.split_whitespace();
			_read!(it; $($t),*)
		}).collect::<Vec<_>>();
		ret
	}};
	($($t:tt),*) => {{
		let line = readln();
		let mut it = line.split_whitespace();
		_read!(it; $($t),*)
	}};
}

macro_rules! _read {
	($it:ident; [char]) => {
		_read!($it; String).chars().collect::<Vec<_>>()
	};
	($it:ident; [u8]) => {
		Vec::from(_read!($it; String).into_bytes())
	};
	($it:ident; usize1) => {
		$it.next().unwrap_or_else(|| panic!("input mismatch")).parse::<usize>().unwrap_or_else(|e| panic!("{}", e)) - 1
	};
	($it:ident; [usize1]) => {
		$it.map(|s| s.parse::<usize>().unwrap_or_else(|e| panic!("{}", e)) - 1).collect::<Vec<_>>()
	};
	($it:ident; [$t:ty]) => {
		$it.map(|s| s.parse::<$t>().unwrap_or_else(|e| panic!("{}", e))).collect::<Vec<_>>()
	};
	($it:ident; $t:ty) => {
		$it.next().unwrap_or_else(|| panic!("input mismatch")).parse::<$t>().unwrap_or_else(|e| panic!("{}", e))
	};
	($it:ident; $($t:tt),+) => {
		($(_read!($it; $t)),*)
	};
}


pub fn main() {
	let _ = ::std::thread::Builder::new().name("run".to_string()).stack_size(32 * 1024 * 1024).spawn(run).unwrap().join();
}

const MOD: u32 = 1_000_000_007;
const INF: i64 = std::i64::MAX/2;

#[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(&self, f: &mut std::fmt::Formatter<'_>) -> std::fmt::Result {
        write!(f, "{}", self.0)
    }
}
*/
 
#[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)
    }
}

fn solve() {
    let (n,c) = read!(usize,usize);
    let a = read!([u32]);
    let b = read!([u32]);
    let mut cof = vec![ModInt(0);2*c];
    cof[0] = ModInt(1);
    for i in 0..n {
        let mut ncof = vec![ModInt(0);2*c];
        for j in 0..c+1 {
            for k in 0..c+1 {
                if j+k>c {
                    break;
                }
                let val = ModInt(a[i]).pow(j as u32)*cof[k];
                ncof[j+k] += val;
            }
        }
        cof = ncof;
    }
    println!("{}", cof[c].0);
}

fn run() {
    solve();
}