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(ll (i) = (0);(i) < (n);++i)
#define REV(i,n) for(ll (i) = (n) - 1;(i) >= (0);--i)
#define ALL(v) v.begin(),v.end()
#define FI first
#define SE second
#define PB push_back
#define MP make_pair
#define INF 1000000000

typedef long long ll;
typedef pair<ll, ll> P;

int dp[66][66][3333];

void chmin(int &a, int b){
    a = min(a, b);
}

int main() {
    ll n, k;cin >> n >> k;
    vector<int> v(n);
    int sum = 0, red_sum = 0, black_sum = 0;
    REP(i, n) {
        cin >> v[i];
        sum += v[i];
    }
    sort(ALL(v), greater<int>());
    vector<int> r(k);
    REP(i, k){
        cin >> r[i];
        red_sum += r[i];
    }
    sort(ALL(r), greater<int>());
    vector<int> b(k);
    REP(i, k){
        cin >> b[i];
        black_sum += b[i];
    }
    sort(ALL(b), greater<int>());

    REP(i, 66)REP(j, 66)REP(k, 3333)dp[i][j][k] = INF;
    dp[0][0][0] = 0;

    REP(red, k+1){
        REP(black, k+1){
            if(red + black >= n)continue;
            REP(red_sum, 3333)if(dp[red][black][red_sum] != INF){
                int id = red + black;
                if(red < k){
                    int nxt_sum = red_sum + v[id];
                    int cost = max(0, v[id] - r[red]);
                    chmin(dp[red+1][black][nxt_sum], dp[red][black][red_sum] + cost);
                }
                if(black < k){
                    int nxt_sum = red_sum;
                    int cost = max(0, v[id] - b[black]);
                    chmin(dp[red][black+1][nxt_sum], dp[red][black][red_sum] + cost);
                }
            }
        }
    }

    int ans = INF;
    REP(i, 66){
        REP(j, 66){
            if(i + j != n)continue;
            REP(k, 3333){
                if(k > red_sum)continue;
                if(sum - k > black_sum)continue;
                chmin(ans, dp[i][j][k]);
            }
        }
    }

    cout << ((ans == INF) ? -1 : ans) << endl;
}