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>
using namespace std;

#define REP(i, n) for(int64 i = 0; i < (n); i++)

using int64 = int64_t;
constexpr int64 INF_LL = (int64)1e18;

template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;};
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;};


int main(void) {
    int64 N, K;
    cin >> N >> K;
    vector<int64> a(N), r(K), b(K);
    vector<int64> sum_a(N+1, 0), sum_r(K+1, 0), sum_b(K+1, 0);
    REP(i, N) { cin >> a[i]; }
    REP(i, K) { cin >> r[i]; }
    REP(i, K) { cin >> b[i]; }
    sort(a.rbegin(), a.rend()); sort(r.rbegin(), r.rend()); sort(b.rbegin(), b.rend());

    REP(i, N) sum_a[i+1] = sum_a[i] + a[i];
    REP(i, K) sum_r[i+1] = sum_r[i] + r[i];
    REP(i, K) sum_b[i+1] = sum_b[i] + b[i];

    int64 all_a = sum_a.back(), all_r = sum_r.back(), all_b = sum_b.back();

    vector<vector<int64>> dp(K+2, vector<int64>(3010, INF_LL));
    dp[0][0] = 0;
    REP(i, N) {
        vector<vector<int64>> dp2(K+2, vector<int64>(3010, INF_LL));
        REP(r_p, min(K+1, i+1)) {
            REP(r_use, 3010) {
                if (r_use + a[i] <= all_r) {
                    chmin(dp2[r_p+1][r_use+a[i]], dp[r_p][r_use] + max((int64)0, a[i]-r[r_p]));
                }
                int64 b_p = i-r_p, b_use = sum_a[i]-r_use;
                if (b_use + a[i] <= all_b) {
                    chmin(dp2[r_p][r_use], dp[r_p][r_use] + max((int64)0, a[i]-b[b_p]));
                }
            }
        }
        swap(dp, dp2);
    }
    int64 res = INF_LL;
    REP(i, K+1) {
        REP(j, 3010) {
//            if (j <= 15) {
//                cout << dp[i][j] << " ";
//            }
            res = min(res, dp[i][j]);
        }
//        cout << endl;
    }
    cout << (res == INF_LL ? -1 : res) << endl;
}
