Codeforces Round #402 (Div. 2) A.Pupils Redistribution 認真讀題

題目：

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

4
5 4 4 4
5 5 4 5

output

1

input

6
1 1 1 1 1 1
5 5 5 5 5 5

output

3

input

1
5
3

output

-1

input

9
3 2 5 5 2 3 3 3 2
4 1 4 1 1 2 4 4 1

output

4

這個題目在比賽中差點沒寫出來，原因麼就是不仔細讀題，直接看樣例，只記得模模糊糊說讓兩組的分數相等。0rzzz，後來打算dp上時發現最後一組樣例不對，於是重新讀題，才發現題讀錯了。***。

題目還是很簡單的，先分別統計兩組中grades1——5的人數，如果每一個分數的人數加起來有奇數，那麼無解。

然後，我們只討論組1比組2多的分數是哪幾個。res+=(cnt1[i]-cnt2[i])/2.至於具體是怎麼換的，我不關心，我只知道當把組1,比達到終點狀態多的換掉，那麼最終結果就出來了。或者組2比組1多的也行。因爲這兩者情況是對稱的。

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);
}

好好讀題呀！