Problem: 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

Solution (in CPP):

Solution:
#include "bits/stdc++.h"

using namespace std;

#define rep(i, n) for (int i = 0; i < n; i++)
#define vi vector<int>
#define all(x) (x).begin(), (x).end()

#define INF 4294967295

unsigned int dp[2][62][62][4000];


signed main()
{
	int nn, kk;
	cin >> nn >> kk;
	vi data(nn);
	rep(i, nn) {
		cin >> data[i];
	}
	vi red(kk), black(kk);
	int RED = 0, BLACK = 0;
	rep(i, kk) {
		cin >> red[i];
		RED += red[i];
	}
	rep(i, kk) {
		cin >> black[i];
		BLACK += black[i];
	}


	sort(all(data));
	reverse(all(data));
	sort(all(red));
	reverse(all(red));
	sort(all(black));
	reverse(all(black));
	vi ruiseki(nn + 1, 0);
	rep(i, nn) {
		ruiseki[i + 1] = ruiseki[i] + data[i];
	}

	//cout << "hoge " << endl;

	//rep(j, kk + 1) {
	//	rep(k, kk + 1) {
	//		rep(l, RED + 1) {
	//			dp[0][j][k][l] = INF;
	//		}
	//	}
	//}
	memset(dp[0], -1, sizeof(dp[0]));

	dp[0][0][0][0] = 0;
	

	//cout << "hoge " << endl;

	int ura = 0;
	int omote = 1;
	rep(i, data.size()) {

		int rr = ruiseki[i];
		int dd = data[i];
		//rep(j, kk + 1) {
		//	rep(k, kk + 1) {
		//		rep(l, RED + 1) {
		//			dp[omote][j][k][l] = INF;
		//		}
		//	}
		//}
		memset(dp[omote], -1, sizeof(dp[omote]));
		rep(j, kk + 1) {
			
			rep(k, kk + 1) {
				for (int l = 0; l <= rr && l + dd <= RED; l++)
				{
					if (dp[ura][j][k][l] ==  INF || j == kk)continue;
						dp[omote][j + 1][k][l + dd] =
							min(dp[omote][j + 1][k][l + dd],
								dp[ura][j][k][l] + max(0, dd - red[j]));
					
				}
				for (int l = rr; 0 <= l && rr - l + dd <= BLACK; l--)
				{
					if (dp[ura][j][k][l] == INF || k == kk)continue;
						dp[omote][j][k + 1][l] =
							min(dp[omote][j][k + 1][l],
								dp[ura][j][k][l] + max(0, dd - black[k]));
					
				}
				
			}
		}
		swap(omote, ura);
	
	}
	unsigned int ans = INF;
	rep(j, kk + 1) {
		rep(k, kk + 1) {
			rep(l, RED + 1) {
				
				ans = min(ans, dp[ura][j][k][l]);
			}
		}
	}
	if (ans == INF) {
		cout << -1 << endl;
	}
	else {

		cout << ans << endl;
	}
	return 0;
}