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
//
// Created by yamunaku on 2019/10/22.
//

#include <bits/stdc++.h>

using namespace std;

#define rep(i, n) for(int i = 0; i < (n); i++)
#define repl(i, l, r) for(int i = (l); i < (r); i++)
#define per(i, n) for(int i = ((n)-1); i >= 0; i--)
#define perl(i, l, r) for(int i = ((r)-1); i >= (l); i--)
#define all(x) (x).begin(),(x).end()
#define MOD9 998244353
#define MOD1 1000000007
#define IINF 1000000000
#define LINF 1000000000000000000
#define SP <<" "<<
#define CYES cout<<"Yes"<<endl
#define CNO cout<<"No"<<endl
#define CFS cin.tie(0);ios::sync_with_stdio(false)
#define CST(x) cout<<fixed<<setprecision(x)

typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef vector<vector<int>> mti;
typedef vector<ll> vl;
typedef vector<vector<ll>> mtl;

int main(){
    int n, k;
    cin >> n >> k;
    vi a(n);
    int s = 0;
    rep(i, n) cin >> a[i], s += a[i];
    sort(all(a), greater<int>());
    vi r(k), b(k);
    int rs = 0, bs = 0;
    rep(i, k) cin >> r[i], rs += r[i];
    sort(all(r), greater<int>());
    rep(i, k) cin >> b[i], bs += b[i];
    sort(all(b), greater<int>());
    vector<mti> dp(k + 1, mti(rs + 1, vi(2, 10000)));
    dp[0][0][0] = 0;
    rep(i, n){
        perl(j, 1, k + 1){
            per(x, rs + 1){
                int tmp = dp[j][x][0];
                if(x >= a[i]){
                    dp[j][x][0] = min(dp[j - 1][x - a[i]][0], dp[j - 1][x - a[i]][1]) + max(0, a[i] - r[j - 1]);
                    dp[j][x][0] = min(10000, dp[j][x][0]);
                }else{
                    dp[j][x][0] = 10000;
                }
                if(0 <= i-j && i-j < k){
                    dp[j][x][1] = min(tmp, dp[j][x][1]) + max(0, a[i] - b[i - j]);
                    dp[j][x][1] = min(10000, dp[j][x][1]);
                }else{
                    dp[j][x][1] = 10000;
                }
            }
        }
        if(i < k){
            per(x, rs + 1){
                dp[0][x][1] = min(dp[0][x][0], dp[0][x][1]) + max(0, a[i] - b[i]);
                dp[0][x][1] = min(10000, dp[0][x][1]);
            }
        }else{
            per(x, rs + 1){
                dp[0][x][1] = 10000;
            }
        }
        dp[0][0][0] = 10000;
    }
    int ans = 10000;
    rep(j, k + 1){
        rep(x, rs + 1){
            if(x <= rs && s - x <= bs){
                ans = min(ans, min(dp[j][x][0], dp[j][x][1]));
            }
        }
    }
    if(ans == 10000) cout << -1 << endl;
    else cout << ans << endl;
    return 0;
}

