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 ll long long
#define maxn 100005 /*rem*/
#define mod 998244353
#define db double
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define pi pair<int, int>
#define fi first
#define se second
using namespace std;
ll ksm(ll a, ll b) {
   if (!b) return 1;
   ll ns = ksm(a, b >> 1);
   ns = ns * ns % mod;
   if (b & 1) ns = ns * a % mod;
   return ns;
}
int dp[125][65][3505]; // 1 ~ i 分配了多少个红色 总和是若干 方法数
int a[2][65];
int s[2][65];
int p[125];
int n, k;
const int inf = 1e9;
int main() {
	cin >> n >> k;
	for (int i = 1; i <= n; i++)
		scanf("%d", &p[i]);
	sort(p + 1, p + n + 1);
	reverse(p + 1, p + n + 1);
	for (int j = 0; j < 2; j++) {
		for (int q = 1; q <= k; q++) 
			scanf("%d", &a[j][q]);
		sort(a[j] + 1, a[j] + k + 1);
		reverse(a[j] + 1, a[j] + k + 1);
		for (int q = 1; q <= k; q++)
			s[j][q] = s[j][q - 1] + a[j][q];
	}
	for (int i = 0; i < 125; i++)
		for (int j = 0; j < 65; j++)
			for (int k = 0; k < 3505; k++)
				dp[i][j][k] = inf;
	for (int j = n + 1; j >= 1; j--) {
		int ss = 0;
		for (int l = 1; l < j; l++) ss += p[l];
		for (int m = 0; m <= k; m++)
			for(int u = 0; u <= 50 * m; u++) {
				int d[2] = {m, j - 1 - m};
				int q[2] = {u, ss - u};
				if (d[0] > k || d[1] > k || d[0] < 0 || d[1] < 0) dp[j][m][u] = inf;
				else {
					if (j == n + 1) {
						if (q[0] > s[0][k] || q[1] > s[1][k]) dp[j][m][u] = inf;
						else {
						//	cout << "!!!" << u << endl;
							dp[j][m][u] = 0;
						}
					}
					else {
						dp[j][m][u] = inf;
						for (int tr = 0; tr < 2; tr++) {
							int cp = max(0, p[j] - a[tr][d[tr] + 1]);
							int em = m, eu = u;
							if (tr == 0) em++, eu += p[j];
							dp[j][m][u] = min(dp[j][m][u], dp[j + 1][em][eu] + cp);
						}
					}
				}
			}
	}
	int ans = dp[1][0][0];
	if (ans > 1e8) cout << -1 << endl;
	else cout << ans << endl;
	return 0;
}

