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# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Linq;
class E { static void Main() { new K(); } }
class K
{
	int[] G() { return Console.ReadLine().Split().Select(_ => int.Parse(_)).ToArray(); }
	long Mod = 1000000007;
	public K()
	{
		var I = G();
		N = I[0]; C = I[1];
		A = G();
		B = G();
		memo = new long[N, C + 1];
		used = new bool[N, C + 1];
		pows = new long[401, C + 1];
		for (var x = 0; x <= 400; x++) pows[x, 0] = 1;
		for (var x = 0; x <= 400; x++) for (var c = 1; c <= C; c++) pows[x, c] = (pows[x, c - 1] * x) % Mod;
		powsum = new long[402, C + 1];
		for (var x = 1; x < 402; x++) for (var c = 0; c <= C; c++) powsum[x, c] = (powsum[x - 1, c] + pows[x - 1, c]) % Mod;
		Console.WriteLine(Calc(0, C));
	}
	int N, C;
	int[] A, B;
	long[,] memo, pows, powsum;
	bool[,] used;
	long Calc(int n, int s)
	{
		if (!used[n, s])
		{
			used[n, s] = true;
			var ans = 0L;
			if (n == N - 1) ans = (powsum[B[n] + 1, s] - powsum[A[n], s]) % Mod;
			else for (var t = 0; t <= s; t++) ans = (ans + ((powsum[B[n] + 1, t] - powsum[A[n], t]) % Mod) * Calc(n + 1, s - t) % Mod) % Mod;
			if (ans < 0) ans += Mod;
			memo[n, s] = ans;
		}
		return memo[n, s];
	}
}