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 <cstdio>
#include <iostream>
#include <vector>
#include <map>
#include <algorithm>

using namespace std;

typedef long long ll;

#define SIZE 500
#define INF 1000000000

int a[SIZE];
int r[SIZE], b[SIZE];

int sumA, sumR, sumB;

int dp[121][61][50*60+1];

int main() {
  int N, K;

  scanf("%d%d", &N, &K);

  for(int i=0; i<N; i++) {
    scanf("%d", a+i);
    sumA += a[i];
  }

  for(int i=0; i<K; i++) {
    scanf("%d", r+i);
    sumR += r[i];
  }

  for(int i=0; i<K; i++) {
    scanf("%d", b+i);
    sumB += b[i];
  }

  for(int i=0; i<=N; i++) {
    for(int j=0; j<=K; j++) {
      for(int k=0; k<=sumR; k++) {
        dp[i][j][k] = INF;
      }
    }
  }

  sort(a, a+N, greater<int>());
  sort(b, b+K, greater<int>());
  sort(r, r+K, greater<int>());

  dp[0][0][0] = 0;

  for(int i=0; i<N; i++) {
    for(int j=0; j<=K; j++) {
      for(int k=0; k<sumR; k++) {

        if(j < K && k + a[i] <= sumR) {
          dp[i+1][j+1][k+a[i]] = min( dp[i+1][j+1][k+a[i]], dp[i][j][k] + max(a[i] - r[j], 0));
        }

        dp[i+1][j][k] = min(dp[i+1][j][k], dp[i][j][k] + max(0, a[i] - b[i-j]));
      }
    }
  }

  int ans = INF;

  for(int j=0; j<=K; j++) {
    for(int k=0; k<=sumR; k++) {
      if(N - j <= K && sumA - k <= sumB)
        ans = min(ans, dp[N][j][k]);
    }
  }

  if (ans == INF)
    cout << -1 << endl;
  else
    cout << ans << endl;

  return 0;
}