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<vector>
#include<algorithm>
#include<utility>
using namespace std;
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define int long long
typedef long long ll;
void amin(int &a, const int &b){
    a = min(a, b);
}
signed main(){
    int N, K;
    cin >> N >> K;
    vector<int> A(N);
    vector<int> R(K);
    vector<int> B(K);
    for(int i=0;i<N;i++){
        cin >> A[i];
    }
    for(int i=0;i<K;i++){
        cin >> R[i];
    }
    for(int i=0;i<K;i++){
        cin >> B[i];
    }

    sort(A.rbegin(), A.rend());
    sort(R.rbegin(), R.rend());
    sort(B.rbegin(), B.rend());
    int INF = (int)1 << 50;
    vector<vector<vector<int> > > dp(R.size() + 1, 
    vector<vector<int> >(B.size() + 1, vector<int>(6005, INF)));
    int offset = 3002;
    dp[0][0][offset] = 0;

    vector<int> SA(N + 1, 0);
    vector<int> SR(N + K, 0);
    vector<int> SB(N + K, 0);

    for(int i=0;i<N;i++){
        SA[i + 1] = SA[i] + A[i];
    }

    for(int i=0;i<K;i++){
        SR[i + 1] = SR[i] + R[i];
        SB[i + 1] = SB[i] + B[i];
    }
    for(int i=0;i<= R.size(); i++){
        for(int j=0;j<= B.size(); j++){
            if(i + j < N){
                for(int k = -3000; k<=3000;k++){
                   

                    if(i + 1 <= R.size()){
                        //if(k + R[i] - A[i + j] >= 0){
                            amin(dp[i + 1][j][k + R[i] - A[i + j] + offset], dp[i][j][k + offset] + max((int)0, A[i + j] - R[i]));
                        //}
                    }
                    if(j + 1 <= B.size()){
                        //if(b + B[i] - A[i + j] >= 0){
                            amin(dp[i][j + 1][k + offset], dp[i][j][k + offset] + max((int)0, A[i + j] - B[j]));
                        //}
                    }
                }
            }
        }
    }
    int res = INF;
    for(int i=0;i<=N;i++){
        int j = N - i;
        if(i <= R.size() && j <= B.size()){
            
            for(int k = -(SR[R.size()] - SR[i]) ; k <= 3000; k++){
                int b = SR[i] + SB[j] - SA[i + j] - k;
                if(b >= -(SB[B.size()] - SB[j])){
                    if(dp[i][j][k + offset] < INF){
                        //cerr << SA[i + j]<<endl;
                        //cerr << i <<" " << j << " " << k << " " << b << endl;
                    }
                    amin(res, dp[i][j][k + offset]);
                }
            }   
        }
    }
    if(res == INF)res = -1;
    cout << res << endl;

    return 0;
}
