/*一定度清楚題意，是小領導要求a，b兩個小組具有相同多的value的學生*/#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int a[55], b[55];int main(){    int n, num;    while(scanf("%d", &n) != EOF){        memset(a, 0, sizeof(a));        memset(b, 0, sizeof(b));        for(int i = 0; i < n; i++){            scanf("%d", &num);            a[num]++;//這種題獨特的技術方法 第二次碰到        }        for(int j = 0; j < n; j++){            scanf("%d", &num);            b[num]++;        }        int ans = 0;        for(int i = 1; i <= 5; i++){            if((a[i] + b[i]) % 2) {cout << "-1" << endl;return 0;}            if((a[i] + b[i]) == 0) continue;            if(((a[i] + b[i]) != 0) && a[i] != b[i]){                ans += abs(a[i] - (a[i] + b[i])/2);            }        }        cout << ans/2 << endl;//因為 共加了兩次    }    return 0;}