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;
int N,K;
int A[66],R[66],B[66];
int sA[66],sR[66],sB[66];
int dp[61][61][6001];
const int off=3000;
inline int chmin(int&a,int b)
{
    if(a>b)a=b;
    return a;
}
int main()
{
    cin>>N>>K;
    for(int i=0;i<N;i++)cin>>A[i];
    sort(A,A+N);
    reverse(A,A+N);
    for(int i=0;i<K;i++)cin>>R[i];
    sort(R,R+K);
    reverse(R,R+K);
    for(int i=0;i<K;i++)cin>>B[i];
    sort(B,B+K);
    reverse(B,B+K);
    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];
    for(int i=0;i<K;i++)sB[i+1]=sB[i]+B[i];
    for(int i=0;i<=N;i++)for(int j=0;j<=K;j++)for(int k=-off;k<=off;k++)
    {
        dp[i][j][k+off]=1e9;
    }
    dp[0][0][off]=0;
    for(int i=0;i<N;i++)
    {
        for(int j=0;j<=K;j++)
        {
            for(int k=-off;k<=off;k++)
            {
                if(dp[i][j][k+off]==1000000000)continue;
                if(j<K)
                {
                    chmin(dp[i+1][j+1][k+off+R[j]-A[i]],dp[i][j][k+off]+(A[i]>R[j]?A[i]-R[j]:0));
                }
                int id=i-j;
                if(0<=id&&id<K)
                {
                    chmin(dp[i+1][j][k+off],dp[i][j][k+off]+(A[i]>B[id]?A[i]-B[id]:0));
                }
            }
        }
    }
    int ans=1e9;
    for(int i=0;i<=K;i++)
    {
        for(int j=-off;j<=off;j++)
        {
            int rrest=j+sR[K]-sR[i];
            int brest=-sA[N]+sR[K]+sB[K]-rrest;
            //if(dp[N][i][j+off]<1e9)cout<<"DBG : "<<rrest<<" "<<brest<<endl;
            if(0<=rrest&&0<=brest)
            {
                ans=min(ans,dp[N][i][j+off]);
            }
        }
    }
    cout<<(ans<1e9?ans:-1)<<endl;
}