codeforces 779A

A. Pupils Redistribution

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

In Berland each high school student is characterized by academic performance — integer value between 1 and 5.

In high school 0xFF there are two groups of pupils: the group A and the group B. Each group consists of exactly n students. An academic performance of each student is known — integer value between 1 and 5.

The school director wants to redistribute students between groups so that each of the two groups has the same number of students whose academic performance is equal to 1, the same number of students whose academic performance is 2 and so on. In other words, the purpose of the school director is to change the composition of groups, so that for each value of academic performance the numbers of students in both groups are equal.

To achieve this, there is a plan to produce a series of exchanges of students between groups. During the single exchange the director selects one student from the class A and one student of class B. After that, they both change their groups.

Print the least number of exchanges, in order to achieve the desired equal numbers of students for each academic performance.

Input

The first line of the input contains integer number n (1 ≤ n ≤ 100) — number of students in both groups.

The second line contains sequence of integer numbers a1, a2, ..., an (1 ≤ ai ≤ 5), where ai is academic performance of the i-th student of the group A.

The third line contains sequence of integer numbers b1, b2, ..., bn (1 ≤ bi ≤ 5), where bi is academic performance of the i-th student of the group B.

Output

Print the required minimum number of exchanges or -1, if the desired distribution of students can not be obtained.

Examples

input

45 4 4 45 5 4 5

output

1

input

61 1 1 1 1 15 5 5 5 5 5

output

3

input

153

output

-1

input

93 2 5 5 2 3 3 3 24 1 4 1 1 2 4 4 1

output

4

题意：给你两个序列，求使得两个序列中相同元素个数相同的最小交换次数。

分析：将两个序列中每个元素的个数分别统计出来，用两个序列中相同元素的个数绝对值之差的二分之一之和求出来，最后再取一半就可以得到正确答案。

//注意，个数之和为奇数时不能再分，这时输出-1即可。

代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn=103;
int a[maxn],b[maxn];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        int sum=0;
        int x;
        int y;
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&x);
            a[x]++;
        }
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&y);
            b[y]++;
        }
        int flag=0;
        for(int i=1;i<=5;i++)
        {
            if((a[i]+b[i])==0) continue;
            if((a[i]+b[i])%2)
            {
                flag=1;
            }
            int t=abs(a[i]-b[i]);
            sum+=t/2;
        }

        if(flag==1) printf("-1\n");
        else printf("%d\n",sum/2);
    }
    return 0;
}