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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
 
#define INF 10000000000
#define M 1000000007

#define min(a,b) (((a)<(b))?(a):(b))
#define max(a,b) (((a)>(b))?(a):(b))

int compare_long(const void *a, const void *b) {
	return *(long*)a - *(long*)b;
}

long combi(int n, int r) {
	int numerator[r];
	int denominator[r];
	int k, p;
	int pivot;
	int offset;
	long ret = 1;
	
	if (n < 0 || r < 0 || r > n) return 0;
	
	if (n - r < r) r = n - r;
	if (r == 0) return 1;
	if (r == 1) return n;
	
	
	
	for(k = 0;k < r;k++) {
		numerator[k] = n - r + k + 1;
		denominator[k] = k + 1;
	}
	
	for(p = 2; p <= r; p++) {
		pivot = denominator[p - 1];
		if(pivot > 1) {
			offset = (n - r) % p;
			for(k = p - 1; k < r; k += p) {
				numerator[k - offset] /= pivot;
				denominator[k] /= pivot;
			}
		}
	}
	
	for(k = 0; k < r; k++) {
		if(numerator[k] > 1) ret = (ret * numerator[k]) % M;
	}
	
	return ret;
}

int main() {
	long N, C;
	long i, j, k;
	long ans;
	long A[401], B[401];
	long com;
	long p;
	long sum = 0;
	
	scanf("%ld %ld", &N, &C);
	for(i=1;i<=C;i++) sum += i;
	for(i=1;i<=N;i++) {
		scanf("%ld", &A[i]);
	}
	for(i=1;i<=N;i++) {
		scanf("%ld", &B[i]);
		if(A[i] != B[i]) {
			printf("-1\n");
			return 0;
		}
	}
	
	com = combi(C+N-1, N-1);
	ans = 0;
	for(i=1;i<=N;i++) {
		for(j=0;j<sum;j++) {
			ans = (ans + (long)pow(A[i], j)) % M;
		}
		ans = (ans + p) % M;
	}
	
	printf("%ld\n", ans);
	
	return 0;
}