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 <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define fi first
#define se second
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
// head

int dp[66][3010][66];
int n,k,a[66],b[66],c[66],sb,sc;
int main() {
	scanf("%d%d",&n,&k);
	memset(dp,0x20,sizeof(dp));
	dp[0][0][0]=0;
	rep(i,0,n) scanf("%d",a+i);
	rep(i,0,k) scanf("%d",b+i),sb+=b[i];
	rep(i,0,k) scanf("%d",c+i),sc+=c[i];
	sort(a,a+n); reverse(a,a+n);
	sort(b,b+k); reverse(b,b+k);
	sort(c,c+k); reverse(c,c+k);
	int ps=0,ans=0x20202020;
	for (int i=0;i<=n;i++) {
		for (int s=0;s<=ps;s++) for (int l=0;l<=i;l++) if (dp[i][s][l]<0x20202020) {
			//printf("%d %d %d %d\n",i,s,l,dp[i][s][l]);
			if (i==n) {
				ans=min(ans,dp[i][s][l]);
				continue;
			}
			// put in 1
			if (l<k&&s+a[i]<=sb) {
				dp[i+1][s+a[i]][l+1]=min(dp[i+1][s+a[i]][l+1],dp[i][s][l]+max(0,a[i]-b[l]));
			}
			if (i-l<k&&ps-s+a[i]<=sc) {
				dp[i+1][s][l]=min(dp[i+1][s][l],dp[i][s][l]+max(0,a[i]-c[i-l]));				
			}
		}
		ps+=a[i];
	}
	if (ans<=(1<<20)) printf("%d\n",ans); else puts("-1");
}
