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 <image> modulo 10^9+7.

Constraints

* 1≦N≦400
* 1≦C≦400
* 1≦A_i≦B_i≦400 (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 <image> modulo 10^9+7.

Examples

Input

2 3
1 1
1 1


Output

4


Input

1 2
1
3


Output

14


Input

2 3
1 1
2 2


Output

66


Input

4 8
3 1 4 1
3 1 4 1


Output

421749


Input

3 100
7 6 5
9 9 9


Output

139123417
#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

const int mod = 1e9 + 7;
int f[405][405], a[405][405], l[405], r[405], n, c, b[405][405], ans;

inline int read()
{
	char ch; int res;
	while (ch = getchar(), ch < '0' || ch > '9');
	res = ch - 48;
	while (ch = getchar(), ch >= '0' && ch <= '9')
	res = res * 10 + ch - 48;
	return res;
}

int main()
{
//	freopen("candy.in", "r", stdin);
//	freopen("candy.out", "w", stdout);
	int i, j, k;
	n = read(); c = read();
	for (i = 1; i <= n; i++) l[i] = read();
	for (i = 1; i <= n; i++) r[i] = read();
	for (i = 1; i <= n; i++)
	for (int x = l[i]; x <= r[i]; x++)
	{
		int y = 1;
		for (j = 0; j <= c; j++)
		{
			a[i][j] = (a[i][j] + y) % mod;
			y = 1ll * y * x % mod;
		}
	}
	f[0][0] = 1;
	for (i = 1; i <= n; i++)
	for (j = 0; j <= c; j++)
	for (k = 0; k <= j; k++)
	f[i][j] = (f[i][j] + 1ll * f[i - 1][k] * a[i][j - k]) % mod;
	cout << f[n][c] << endl;
	fclose(stdin);
	fclose(stdout);
	return 0;
}