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 <vector>
#include <algorithm>
#include <iterator>
#include <array>
#include <cmath>

using namespace std;

using IntType = long long;

const int CHILD_MAX = 401;
const int CANDY_MAX = 401;
const int HASYAGI_MAX = 401;
int childNum;
int candyNum;
int hasyagis[2][CHILD_MAX];
const IntType mod = 1000000007;
#define REP(i,n) FOR(i,0,n)
#define FOR(i,a,b) for(int i = (a); i < (b); i++)

IntType powAccumSum[CANDY_MAX][HASYAGI_MAX];
void calcPowAccumSum() {
	REP(i, HASYAGI_MAX) {
		powAccumSum[0][i] = 1;
	}
	powAccumSum[0][0] = 0;
	FOR(a, 1, CANDY_MAX) { // calc i^a
		REP(i, HASYAGI_MAX) {
			powAccumSum[a][i] = (powAccumSum[a - 1][i] * i) % mod;
		}
	}
	REP(a, CANDY_MAX) { // calc sum_i i^a
		FOR(i, 1, HASYAGI_MAX) {
			powAccumSum[a][i] += powAccumSum[a][i - 1];
			powAccumSum[a][i] %= mod;
		}
	}
}
IntType polyDp[CHILD_MAX][CANDY_MAX];
void calcPolyDp() {
	polyDp[0][0] = 1;
	FOR(childI, 1, childNum + 1) REP(candyI, candyNum + 1) {
		REP(k, candyI + 1) {
			polyDp[childI][candyI] += polyDp[childI - 1][k] * (powAccumSum[candyI-k][hasyagis[1][childI]] - powAccumSum[candyI-k][hasyagis[0][childI] - 1]);
			polyDp[childI][candyI] %= mod;
		}
	}
}
int main() {
	cin >> childNum >> candyNum;
	REP(i, 2) FOR(j, 1, childNum + 1) {
		cin >> hasyagis[i][j];
	}
	calcPowAccumSum();
	calcPolyDp();

	cout << polyDp[childNum][candyNum] << endl;

	return 0;
}