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>
using namespace std;

int dp[61][61][3001];

int main(){
    int N,K;
    cin >> N >> K;
    vector<int> A(N+1),R(K+1),B(K+1);
    int sumA = 0,sumR = 0,sumB = 0;
    for(int i=1;i<=N;i++){
        cin >> A[i];
        sumA += A[i];
    }
    for(int i=1;i<=K;i++){
        cin >> R[i];
        sumR += R[i];
    }
    for(int i=1;i<=K;i++){
        cin >> B[i];
        sumB += B[i];
    }
    sort(A.begin()+1,A.end(),greater<int>());
    sort(R.begin()+1,R.end(),greater<int>());
    sort(B.begin()+1,B.end(),greater<int>());
    for(int i=0;i<=N;i++) for(int j=0;j<=N;j++) for(int k=0;k<=3000;k++) dp[i][j][k] = 1e9;
    dp[0][0][0] = 0;
    for(int i=1;i<=N;i++){
        for(int j=0;j<=i;j++){
            int rid = j,bid = i-j;
            for(int k=0;k<=3000;k++){
                if(1<=rid && rid<=K && k+A[i]<=3000){
                    dp[i][rid][k+A[i]] = min(dp[i][rid][k+A[i]],dp[i-1][rid-1][k]+max(0,A[i]-R[rid]));
                }
                if(1<=bid && bid<=K){
                    dp[i][i-bid][k] = min(dp[i][i-bid][k],dp[i-1][i-bid][k]+max(0,A[i]-B[bid]));
                }
            }
        }
    }
    int inf = 1e9;
    int res = inf;
    for(int j=0;j<=N;j++){
        for(int k=0;k<=3000;k++){
            int useR = k,useB = sumA-k;
            if(useR<=sumR && useB<=sumB) res = min(res,dp[N][j][k]);
        }
    }
    cout << (res!=inf? res:-1) << endl;
}
