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;
typedef long long ll;
typedef vector<ll> vl;
typedef vector<vl> mt;
typedef vector<mt> v3;

const ll INF = 1ll << 58;

ll n,k;
vl a,r,b;
ll sum_a,sum_r,sum_b;
ll ans = INF;

v3 dp;

int main(){
	cin >> n >> k;
	a = vl(n);
	r = b = vl(k);
	for(ll i = 0;i < n;i++) cin >> a[i];
	for(ll i = 0;i < k;i++) cin >> r[i];
	for(ll i = 0;i < k;i++) cin >> b[i];
	sort(a.rbegin(),a.rend());
	sort(r.rbegin(),r.rend());
	sort(b.rbegin(),b.rend());

	sum_a = accumulate(a.begin(),a.end(),0ll);
	sum_r = accumulate(r.begin(),r.end(),0ll);
	sum_b = accumulate(b.begin(),b.end(),0ll);
	dp = v3(k+2,mt(k+2,vl(sum_a*2+1,INF)));
	dp[0][0][0] = 0;
	for(ll i = 0;i <= k;i++){
		for(ll j = 0;j <= k;j++){
			ll t = i+j;
			if(t >= n) continue;
			for(ll l = 0;l <= sum_a;l++){
				if(i < k) dp[i+1][j][l+a[t]] = min(dp[i+1][j][l+a[t]],dp[i][j][l]+max(0ll,a[t]-r[i]));
				if(j < k) dp[i][j+1][l] = min(dp[i][j+1][l],dp[i][j][l]+max(0ll,a[t]-b[j]));
			}
		}
	}

	for(ll i = 0;i <= k;i++){
		for(ll j = 0;j <= k;j++){
			if(i+j == n){
				for(ll l = 0;l <= sum_a*2;l++){
					if(l <= sum_r && sum_a-l <= sum_b){
						ans = min(ans,dp[i][j][l]);
					}
				}
			}
		}
	}
	if(ans == INF){
		cout << -1 << endl;
	}else{
		cout << ans << endl;
	}
}