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;

#define rep(i,n) for(int i=0;i<n;i++)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)

const int mod=1000000007;
int dp[70][70][6100];
void solve(){
    int n,k;cin>>n>>k;
    vector<int> ra(n+1);
    vector<int> rr(k+1);
    vector<int> rb(k+1);
    vector<int> a(n);
    vector<int> r(k),b(k);
    rep(i,n){
        cin>>a[i];
    }
    rep(i,k){
        cin>>r[i];
    }
    rep(i,k){
        cin>>b[i];
    }
    sort(a.begin(),a.end(),greater<int>());
    sort(r.begin(),r.end(),greater<int>());
    sort(b.begin(),b.end(),greater<int>());
    rep(i,n){
        ra[i+1]=ra[i]+a[i];
    }
    rep(i,k){
        rr[i+1]=rr[i]+r[i];
        rb[i+1]=rb[i]+b[i];
    }
    rep(i,n+1){
        rep(j,k+1){
            rep(l,6100){
                dp[i][j][l]=mod;
            }
        }
    }
    dp[0][0][3010]=0;
    rep(i,n){
        rep(j,k+1){
            int rest=i-j;
            if(rest<0||rest>k)continue;
            rep(l,6100){
                if(dp[i][j][l]==mod)continue;
                //take red
                if(j<k){
                    int dif=r[j]-a[i];
                    int cost=max(0,-dif);
                    dp[i+1][j+1][l+dif]=min(dp[i+1][j+1][l+dif],dp[i][j][l]+cost);
                }
                //take blue
                if(rest<k){
                    int dif=b[rest]-a[i];
                    int cost=max(0,-dif);
                    dp[i+1][j][l]=min(dp[i+1][j][l],dp[i][j][l]+cost);
                }
            }
        }
    }
    int ans=mod;
    rep(j,k+1){
        int rest=n-j;
        if(rest<0)continue;
        if(rest>k)continue;
        rep(l,6100){
            int x=l-3010;
            if(x+rr[k]-rr[j]<0)continue;
            int bb=rr[j]-x+rb[rest]-ra[n];
            if(bb+rb[k]-rb[rest]<0)continue;

            ans=min(ans,dp[n][j][l]);

        }
    }
    if(ans==mod)ans=-1;
    cout<<ans<<endl;

}
int main(){
    solve();
    return 0;
}