E: Red Black Balloons

Story

Homura-chan's dream comes true. It means ICPC Asia regional contest 20xx will be held in Sapporo! Homura-chan has been working hard for the preparation. And finally, it's the previous day of the contest. Homura-chan started to stock balloons to be delivered to contestants who get accepted. However, she noticed that there were only two colors of balloons: red and black.

Problem Statement

ICPC Asia regional contest in Sapporo plans to provide N problems to contestants. Homura-chan is a professional of contest preparation, so she already knows how many contestants would get acceptance for each problem (!!), a_i contestants for the i-th problem. You can assume the prediction is perfectly accurate. Ideally, Homura-chan would assign a distinct color of a balloon to each problem respectively. But you know, she has only two colors red and black now. Thus, Homura-chan comes up with the idea to differentiate the size of balloons in K levels, that is, each problem has a balloon with a distinct pair of (color, size).

Homura-chan now has r_i red balloons with size i (1 \leq i \leq K) and b_j black balloons with size j (1 \leq j \leq K). Suppose we assign a pair (c_i, s_i) of a color c_i (red or black) and a size s_i to the i-th problem, for each i. As noted, we have to assign distinct pairs to each problem, more precisely, (c_i, s_i) \neq (c_j, s_j) holds for i \neq j. Moreover, the number of balloons with (c_i, s_i) must be no less than a_i. In the case that there are no such assignments, Homura-chan can change the size of balloons by her magic power. Note that Homura-chan doesn't know magic to change the color of balloons, so it's impossible.

Your task is to write a program computing the minimum number of balloons whose size is changed by Homura-chan's magic to realize an assignment satisfying the above-mentioned conditions. If there is no way to achieve such an assignment even if Homura-chan changes the size of balloons infinitely, output `-1` instead.

Input


N K
a_1 ... a_N
r_1 ... r_K
b_1 ... b_K


Constraints

* 1 \leq N,K \leq 60
* N \leq 2K
* 1 \leq a_i \leq 50 (1 \leq i \leq N)
* 1 \leq r_j,b_j \leq 50 (1 \leq j \leq K)
* Inputs consist only of integers.



Output

Output the minimum number of balloons whose size is changed to achieve an assignment in a line. If there are no ways to achieve assignments, output `-1` instead.

Sample Input 1


3 2
6 5 4
8 1
7 1


Output for Sample Input 1


3

Homura-chan changes the size of three red balloons from 1 to 2. Then she can assign (black,1) to the problem 1, (red,1) to the problem 2, and (red,2) to the problem 3.

Sample Input 2


2 1
50 50
2
3


Output for Sample Input 2


-1

Sample Input 3


4 3
3 10 28 43
40 18 2
26 7 11


Output for Sample Input 3


5





Example

Input

3 2
6 5 4
8 1
7 1


Output

3
#include <iostream>
#include <vector>
#include <algorithm>
#include <tuple>
#include <array>
#include <map>

#define ll long long

#define TEST

using namespace std;
const int INF = 1e9;
int dp[65][65][4000];
void chmin(int& a, int b) {
	a = min(a, b);
}

int main() {
	for (int i = 0; i < 65; i++)
		for (int j = 0; j < 65; j++)
			for (int k = 0; k < 4000; k++)
				dp[i][j][k] = INF;
	dp[0][0][0] = 0;
	int N, K; cin >> N >> K;
	vector<int> a(N), r(K), b(K);
	for (int i = 0; i < N; i++)cin >> a[i];
	for (int i = 0; i < K; i++)cin >> r[i];
	for (int i = 0; i < K; i++)cin >> b[i];
	sort(a.rbegin(), a.rend());
	sort(r.rbegin(), r.rend());
	sort(b.rbegin(), b.rend());
	int sumred = 0, sumblack = 0, sumQ = 0;
	for (int i = 0; i < N; i++)sumQ += a[i];
	for (int i = 0; i < K; i++)sumred += r[i];
	for (int i = 0; i < K; i++)sumblack += b[i];
	for (int Q = 0; Q < N; Q++) {
		for (int i = 0; i <= Q; i++) {
			int j = Q - i;
			for (int k = 0; k < 3005; k++) {
				if (i + 1 <= K)chmin(dp[i + 1][j][k + a[Q]], dp[i][j][k] + max(a[Q] - r[i], 0));
				if (j + 1 <= K)chmin(dp[i][j + 1][k], dp[i][j][k] + max(a[Q] - b[j], 0));
			}
		}
	}
	/*
	for (int i = 0; i < 4; i++) {
		for (int j = 0; j < 4; j++) {
			for (int k = 0; k < 16; k++) {
				cout << i << " " << j << " " << k << " " << dp[i][j][k] << endl;
			}
		}
	}
	*/
	int ans = INF;
	for (int i = 0; i < 65; i++) {
		for (int j = 0; j < 65; j++) {
			for (int k = 0; k < 4000; k++) {
				if (i + j != N)continue;
				if (k > sumred)continue;
				if (sumQ - k > sumblack)continue;
				chmin(ans, dp[i][j][k]);
			}
		}
	}
	if (ans >= INF)ans = -1;
	cout << ans << endl;
	return 0;
}

