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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

////////////////////////////////////////////
///                                      ///
///         Template ver. 1 rapel        ///
/// Fear is Temporary, Regret is Forever ///
///          Must Try and Get AC         ///
///                                      ///
////////////////////////////////////////////
#include <stdio.h>
#include <iostream>
#include <sstream>
#include <iomanip>

#include <algorithm>
#include <stdlib.h>
#include <string.h>
#include <string>
#include <utility>

#include <math.h>
#include <complex>

#include <assert.h>
#include <time.h>
//#include <chrono>
//#include <random>

#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <list>
#include <bitset>

#define FI first
#define SE second
#define MP make_pair
#define PB push_back
#define PF push_front
#define POB pop_back
#define POF pop_front
#define endl '\n'
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;

void desperate_optimization(int precision){
  	ios_base::sync_with_stdio(false);
  	cin.tie(0);
  	cout.tie(0);
  	cout.setf(ios::fixed);
  	cout.setf(ios::showpoint);
  	cout.precision(precision);
}

const int N = 400;
const LL MOD = 1e9 + 7;
LL psum[N + 5][N + 5];
LL val[N + 5];

void pre() {
	for(int i = 1;i <= N;i++) val[i] = 1;
	for(int i = 0;i <= N;i++) {
		for(int j = 1;j <= N;j++) psum[i][j] = (psum[i][j-1] + val[j]) % MOD;
		for(int j = 1;j <= N;j++) val[j] = val[j] * j % MOD;
	}
}

int A[N + 5],B[N + 5];
LL dp[N + 5][N + 5];

LL solve(int idx,int sisa) {
	if(idx == 0) return (psum[sisa][B[idx]] + MOD - psum[sisa][A[idx] - 1]) % MOD;
	LL &res = dp[idx][sisa];
	if(res != -1) return res;
	res = 0;
	for(int i = 0;i <= sisa;i++) {
		res += solve(idx - 1, sisa - i) * (psum[i][B[idx]] + MOD - psum[i][A[idx] - 1]) % MOD;
	}
	return res %= MOD;
}

void hora(int tc) {
	int n,m;
	cin>>n>>m;
	for(int i = 0;i < n;i++) cin>>A[i];
	for(int i = 0;i < n;i++) cin>>B[i];
	memset(dp, -1, sizeof dp);
	cout<<solve(n - 1, m)<<endl;
}

int main(){
	pre();
	desperate_optimization(10);
	int ntc = 1;
	//cin>>ntc;
	for(int tc = 1;tc <= ntc;tc++) hora(tc);
	return 0;
}