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 <bits/stdc++.h>
#define all(vec) vec.begin(), vec.end()
using namespace std;
using ll = long long;
using P = pair<int, int>;
constexpr ll INF = (1LL << 30) - 1LL;
constexpr ll LINF = (1LL << 60) - 1LL;
constexpr ll MOD = 1e9 + 7;
template <typename T> void chmin(T &a, T b) { a = min(a, b); }
template <typename T> void chmax(T &a, T b) { a = max(a, b); }
int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n, m;
    cin >> n >> m;
    vector<int> a(n), r(m), b(m);
    int asum = 0;
    for (int i = 0; i < n; i++) {
        cin >> a[i];
        asum += a[i];
    }
    int rsum = 0, bsum = 0;
    for (int i = 0; i < m; i++) {
        cin >> r[i];
        rsum += r[i];
    }
    for (int i = 0; i < m; i++) {
        cin >> b[i];
        bsum += b[i];
    }
    sort(all(a));
    sort(all(r));
    sort(all(b));
    reverse(all(a));
    reverse(all(r));
    reverse(all(b));
    vector<vector<vector<int>>> dp(
        n + 1, vector<vector<int>>(m + 1, vector<int>(50 * (n + 1), INF)));
    dp[0][0][0] = 0;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j <= m; j++) {
            for (int k = 0; k <= 50 * (i + 1); k++) {
                if (i >= j && i - j < m) {
                    chmin(dp[i + 1][j][k],
                          dp[i][j][k] + max(0, a[i] - b[i - j]));
                }
                if (j > 0 && k >= a[i]) {
                    chmin(dp[i + 1][j][k],
                          dp[i][j - 1][k - a[i]] + max(0, a[i] - r[j - 1]));
                }
            }
        }
    }
    int res = INF;
    for (int j = 0; j <= m; j++) {
        for (int k = 0; k <= 50 * n; k++) {
            if (k <= rsum && asum - k <= bsum) {
                chmin(res, dp[n][j][k]);
            }
        }
    }
    if (res == INF) {
        cout << -1 << endl;
        return 0;
    }
    cout << res << endl;
}
