Codeforces Round #402 (Div. 2) A.Pupils Redistribution 认真读题

题目：

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 1and 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

这个题目在比赛中差点没写出来，原因么就是不仔细读题，直接看样例，只记得模模糊糊说让两组的分数相等。0rzzz，后来打算dp上时发现最后一组样例不对，于是重新读题，才发现题读错了。***。

题目还是很简单的，先分别统计两组中grades1——5的人数，如果每一个分数的人数加起来有奇数，那么无解。

然后，我们只讨论组1比组2多的分数是哪几个。res+=(cnt1[i]-cnt2[i])/2.至于具体是怎么换的，我不关心，我只知道当把组1,比达到终点状态多的换掉，那么最终结果就出来了。或者组2比组1多的也行。因为这两者情况是对称的。

code：

#include<cstdio>
int cnt1[6],cnt2[6];
int main(){
    int n;scanf("%d",&n);
    for(int i=0;i<n;++i){
        int a;scanf("%d",&a);
        cnt1[a]++;
    }
    for(int i=0;i<n;++i){
        int a;scanf("%d",&a);
        cnt2[a]++;
    }
    for(int i=1;i<=5;++i){
        if((cnt1[i]+cnt2[i])&1){
            printf("-1\n");
            return 0;
        }
    }
    int res=0;
    for(int i=1;i<=5;++i){
        if(cnt1[i]>cnt2[i]){
            res+=(cnt1[i]-cnt2[i])/2;
        }
    }
    printf("%d\n",res);
}

好好读题呀！