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<iostream>
#include<vector>
#include<algorithm>
using namespace std;

int main() {

  // input
  int n, K; cin >> n >> K;
  int sum_b = 0;
  int sum_r = 0;

  vector<int> a(n);
  vector<int> acc(n + 1, 0);
  for(int i = 0; i < n; i++){
    cin >> a[i];
    acc[i + 1] = a[i];
    acc[i + 1] += acc[i];
  }
  vector<int> r(K);
  for(int i = 0; i < K; i++){
    cin >> r[i];
    sum_r += r[i];
  }
  vector<int> b(K);
  for(int i = 0; i < K; i++){
    cin >> b[i];
    sum_b += b[i];
  }
  sort(a.rbegin(), a.rend());
  sort(b.rbegin(), b.rend());
  sort(r.rbegin(), r.rend());

  int inf = 10101010;
  vector<vector<vector<int>>> dp(n + 1, vector<vector<int>> (K + 1, vector<int> (sum_r + 1, inf)));
  dp[0][0][0] = 0;

  for(int i = 0; i < n; i++) {
    for(int j = 0; j < K; j++) {
      for(int k = 0; k < sum_r; k++) {
        if(dp[i][j][k] == inf) continue;


        int l = i - j;
        int used_r = k;
        int used_b = acc[i] - used_r;

        // use red
        int nxt_used_r = used_r + a[i];
        int nxt_cost = max(0, a[i] - r[j]);
        if(nxt_used_r <= sum_r) dp[i + 1][j + 1][nxt_used_r] = min(dp[i + 1][j + 1][nxt_used_r], dp[i][j][k] + nxt_cost);


        // use blue
        int nxt_used_b = used_b + a[i];
        if(l >= K) continue;
        nxt_cost = max(0, a[i] - b[l]);
        if(nxt_used_b <= sum_b) dp[i + 1][j][used_r] = min(dp[i + 1][j][used_r], dp[i][j][k] + nxt_cost);

      }
    }
  }

  int ans = inf;
  for(int j = 0; j <= K; j++)
    for(int k = 0; k <= sum_r; k++)
      ans = min(ans, dp[n][j][k]);

  if(ans == inf) ans = -1;

  cout << ans << endl;
  return 0;
}

