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: Wrong Answer)

#include <cstdio>
#include <iostream>
#include <cmath>
#include <cstring>
#include <sstream>
#include <algorithm>
#include <cstdlib>
#include <map>
#include <queue>
#include <utility>
#include <vector>
#include <set>
#include <memory.h>
#include <iomanip>
#include <bitset>
#include <list>
#include <stack>
#include <deque>
#include <numeric>

using namespace std;

#define mod 1000000007

int n, c;
int a[410];
int b[410];

// powsum[i][j] = 0^i + 1^i + ... + (j - 1)^i
long long int powsum[410][410] = {};

long long int getpow(int x, int a)
{
	if(a == 0) return 1;
	if(a == 1) return x;
	long long int tmpres = getpow(x, a / 2);
	long long int ans = (tmpres * tmpres) % mod;
	if(a % 2 == 1) ans = (ans * x) % mod;
	return ans;
}

void getpowsum()
{
	for(int i = 0; i <= 400; i++){
		for(int j = 1; j <= 401; j++){
			powsum[i][j] = (powsum[i][j - 1] + getpow(j - 1, i)) % mod;
		}
	}
}

long long int result[401][401] = {};

long long int solve(int now, int rest)
{
	if(result[now][rest] > 0) return result[now][rest];
	long long int ans = 0;
	if(now == n - 1){
		ans = powsum[rest][b[now] + 1] - powsum[rest][a[now]];
	} else {
		for(int i = 0; i <= rest; i++){
			long long int tmpres = powsum[i][b[now] + 1] - powsum[i][a[now]];
			tmpres = (tmpres * solve(now + 1, rest - i)) % mod;
			ans = (ans + tmpres) % mod;
		}
	}
	return result[now][rest] = ans;
}

int main()
{
	cin >> n >> c;
	for(int i = 0; i < n; i++){
		cin >> a[i];
	}
	for(int i = 0; i < n; i++){
		cin >> b[i];
	}
	getpowsum();
	cout << solve(0, c) << endl;
}