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;

using vi=vector<int>;
using vvi=vector<vi>;
using vvvi=vector<vvi>;

const int INF=1e9;
const int SIZE=6002;
const int OFFSET=3001;

void cmin(int &lhs,int rhs){
    lhs=min(lhs,rhs);
}
int main(){
    int N,K;
    cin>>N>>K;
    vector<int> a(N),r(K),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.begin(),a.end());
    sort(r.begin(),r.end());
    sort(b.begin(),b.end());
    vi rsum(K+1),bsum(K+1);
    for(int i=0;i<K;i++) rsum[i+1]=rsum[i]+r[i];
    for(int i=0;i<K;i++) bsum[i+1]=bsum[i]+b[i];
    vi asum(N+1);
    for(int i=0;i<=N;i++) asum[i+1]=asum[i]+a[i];
    auto sumR=[&](int l,int r){
        return rsum[r]-rsum[l];
    };
    auto sumB=[&](int l,int r){
        return bsum[r]-bsum[l];
    };
    auto sumA=[&](int l,int r){
        return asum[r]-asum[l];
    };

    vvvi dp(K+1,vvi(K+1,vi(SIZE+1,INF)));
    dp[0][0][OFFSET]=0;
    for(int i=0;i<=K;i++){
        for(int j=0;j<=K;j++){
            for(int X=0;X<=SIZE;X++){
                int k=N-(i+j)-1;
                if(k==-1) continue;
                int Y=sumR(K-i,K)+sumB(K-j,K)-sumA(N-(i+j),N)-(X-OFFSET);
                if(i==1 && j==1 && X==OFFSET+3){
                    
                }
                if(i+1<=K){
                    int toX=X-a[k]+r[K-i-1];
                    if(0<=toX && toX<=SIZE){
                        cmin(dp[i+1][j][toX],dp[i][j][X]+max(0,X-toX));
                    }
                }
                if(j+1<=K){
                    int toY=Y-a[k]+b[K-j-1];
                    cmin(dp[i][j+1][X]  ,dp[i][j][X]+max(0,Y-toY));
                }
            }
        }
    }
    int res=INF;
    for(int i=0;i<=K;i++){
        for(int j=0;j<=K;j++){
            for(int X=0;X<=SIZE;X++){
                int Y=sumR(K-i,K)+sumB(K-j,K)-sumA(N-(i+j),N)-(X-OFFSET);
                if(sumR(0,K-i)+X-OFFSET>=0 && sumB(0,K-j)+Y>=0 && i+j==N){
                    cmin(res,dp[i][j][X]);
                }
            }
        }
    }
    // cerr<<dp[0][1][OFFSET]<<endl;
    // cerr<<dp[1][1][OFFSET+3]<<endl;
    // cout<<dp[2][1][OFFSET]<<endl;
    cout<<(res==INF ? -1 : res)<<endl;
    
    return 0;
}
