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 <algorithm>
#include <vector>
#include <utility>
#define llint long long
#define inf 1000000000

using namespace std;
typedef pair<llint, llint> P;

int n, m;
int a[65], r[65], b[65];
int suma[65], sumr[65], sumb[65];
int dp[65][65][6005];
const int Zero = 3002;

int main(void)
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	cin >> n >> m;
	for(int i = 1; i <= n; i++) cin >> a[i];
	for(int i = 1; i <= m; i++) cin >> r[i];
	for(int i = 1; i <= m; i++) cin >> b[i];
	sort(a+1, a+n+1), sort(r+1, r+m+1), sort(b+1, b+m+1);
	reverse(a+1, a+n+1), reverse(r+1, r+m+1), reverse(b+1, b+m+1);
	
	for(int i = 1; i <= n; i++) suma[i] = suma[i-1] + a[i];
	for(int i = 1; i <= m; i++) sumr[i] = sumr[i-1] + r[i];
	for(int i = 1; i <= m; i++) sumb[i] = sumb[i-1] + b[i];
	
	for(int i = 0; i <= n; i++){
		for(int j = 0; j <= m; j++){
			for(int k = -3000; k <= 3000; k++){
				dp[i][j][Zero+k] = inf;
			}
		}
	}
	dp[0][0][Zero+sumr[m]] = 0;
	
	for(int i = 0; i < n; i++){
		for(int j = 0; j <= m; j++){
			if(i < j) continue;
			for(int k = -3000; k <= 3000; k++){
				int nk = k-a[i+1];
				if(nk >= -3000 && nk <= 3000 && j+1 <= m) dp[i+1][j+1][Zero+nk] = min(dp[i+1][j+1][Zero+nk], dp[i][j][Zero+k] + max(0, a[i+1]-r[j+1]));
				if(i-j+1 <= m) dp[i+1][j][Zero+k] = min(dp[i+1][j][Zero+k], dp[i][j][Zero+k] + max(0, a[i+1]-b[i-j+1]));
			}
		}
	}
	
	/*for(int i = 0; i <= n; i++){
		for(int j = 0; j <= m; j++){
			for(int k = -3000; k <= 3000; k++){
				cout << dp[i][j][Zero+k] << " ";
			}
			cout << endl;
		}
		cout << endl;
	}*/
	
	int ans = inf;
	for(int j = 0; j <= m; j++){
		for(int k = 0; k <= 3000; k++){
			int remR = k, usedR = sumr[m] - remR;
			int usedB = suma[n] - usedR, remB = sumb[m] - usedB;
			//cout << remR << " " << remB << " " << dp[n][j][Zero+k] << endl;
			if(remR < 0 || remB < 0) continue;
			ans = min(ans, dp[n][j][Zero+k]);
		}
	}
	if(ans >= inf) ans = -1;
	cout << ans << endl;
	
	return 0;
}

