#include<bits/stdc++.h>
#define N 65
using namespace std;
int n,K,ans,a[N],r[N],b[N],s0[N],s1[N],s2[N],dp[N][N][N*N];
int main() 
{
	scanf("%d%d",&n,&K);
	for (int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	for (int i=1;i<=K;i++)
		scanf("%d",&r[i]);
	for (int i=1;i<=K;i++)
		scanf("%d",&b[i]);
	sort(a+1,a+n+1,greater<int>());
	sort(r+1,r+K+1,greater<int>());
	sort(b+1,b+K+1,greater<int>());
	for (int i=1;i<=n;i++)
		s0[i]=s0[i-1]+a[i];
	for (int i=1;i<=K;i++)
	{
		s1[i]=s1[i-1]+r[i];
		s2[i]=s2[i-1]+b[i];
	}
	/*if (s0[n]>s1[K]+s2[K])
	{
		puts("-1");
		return 0;
	}*/
	for (int i=0;i<=K;i++)
		for (int j=0;j<=K;j++)
			for (int k=0;k<=s0[n];k++)
				dp[i][j][k]=-12345679;
	dp[0][0][0]=0;
	ans=-1;
	for (int i=0;i<=K;i++)
		for (int j=0;j<=min(K,n-i);j++)
			for (int k=0;k<=s0[i+j];k++)
			{
				if (i&&k>=a[i+j])
					dp[i][j][k]=max(dp[i][j][k],dp[i-1][j][k-a[i+j]]+min(a[i+j],r[i]));
				if (j)
					dp[i][j][k]=max(dp[i][j][k],dp[i][j-1][k]+min(a[i+j],b[j]));
				if (i+j==n&&s1[K]>=k&&s2[K]>=s0[n]-k)
					ans=max(ans,dp[i][j][k]);
			}
	if (s0[n]>s1[K]+s2[K])
		printf("%d\n",-1); else
		printf("%d\n",(ans==-1)?-1:(s0[n]-ans));
	return 0;
}

### Prompt

Please create a solution in cpp to the following 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

### Response

```cpp
#include<bits/stdc++.h>
#define N 65
using namespace std;
int n,K,ans,a[N],r[N],b[N],s0[N],s1[N],s2[N],dp[N][N][N*N];
int main() 
{
	scanf("%d%d",&n,&K);
	for (int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	for (int i=1;i<=K;i++)
		scanf("%d",&r[i]);
	for (int i=1;i<=K;i++)
		scanf("%d",&b[i]);
	sort(a+1,a+n+1,greater<int>());
	sort(r+1,r+K+1,greater<int>());
	sort(b+1,b+K+1,greater<int>());
	for (int i=1;i<=n;i++)
		s0[i]=s0[i-1]+a[i];
	for (int i=1;i<=K;i++)
	{
		s1[i]=s1[i-1]+r[i];
		s2[i]=s2[i-1]+b[i];
	}
	/*if (s0[n]>s1[K]+s2[K])
	{
		puts("-1");
		return 0;
	}*/
	for (int i=0;i<=K;i++)
		for (int j=0;j<=K;j++)
			for (int k=0;k<=s0[n];k++)
				dp[i][j][k]=-12345679;
	dp[0][0][0]=0;
	ans=-1;
	for (int i=0;i<=K;i++)
		for (int j=0;j<=min(K,n-i);j++)
			for (int k=0;k<=s0[i+j];k++)
			{
				if (i&&k>=a[i+j])
					dp[i][j][k]=max(dp[i][j][k],dp[i-1][j][k-a[i+j]]+min(a[i+j],r[i]));
				if (j)
					dp[i][j][k]=max(dp[i][j][k],dp[i][j-1][k]+min(a[i+j],b[j]));
				if (i+j==n&&s1[K]>=k&&s2[K]>=s0[n]-k)
					ans=max(ans,dp[i][j][k]);
			}
	if (s0[n]>s1[K]+s2[K])
		printf("%d\n",-1); else
		printf("%d\n",(ans==-1)?-1:(s0[n]-ans));
	return 0;
}

```