Codeforces Round #199 (Div. 2) -- A. Xenia and Divisors (思路)

A. Xenia and Divisors

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Xenia the mathematician has a sequence consisting of n (n is divisible by 3) positive integers, each of them is at most 7. She wants to split the sequence into groups of three so that for each group of three a, b, c the following conditions held:

• a < b < c;
• a divides b,b divides c.

Naturally, Xenia wants each element of the sequence to belong to exactly one group of three. Thus, if the required partition exists, then it has groups of three.

Help Xenia, find the required partition or else say that it doesn't exist.

Input

The first line contains integer n (3 ≤ n ≤ 99999) — the number of elements in the sequence. The next line containsn positive integers, each of them is at most 7.

It is guaranteed that n is divisible by 3.

Output

If the required partition exists, print groups of three. Print each group as values of the elements it contains. You should print values in increasing order. Separate the groups and integers in groups by whitespaces. If there are multiple solutions, you can print any of them.

If there is no solution, print -1.

Examples

Input

61 1 1 2 2 2

Output

-1

Input

62 2 1 1 4 6

Output

1 2 41 2 6大体题意：给你n（n 一定是3的倍数）个数，1 <= ai <= 7,问是否能挑出n/3组组合 a，b，c 使得 a < b < c;  a divides b, b divides c. 
不可以的话输出-1.思路：想一想便可以知道  组合只有三种情况：1 2 41 2 61 3 6.那么讨论各个数的关系即可了！写这篇博客目的在于 讨论问题 要全面吧，自己认为的没用的条件  便可能是关键问题！！这种题 就在于逻辑顺序了!
#include<cstdio>#include<cstring>#include<cctype>#include<cstdlib>#include<cmath>#include<iostream>#include<sstream>#include<iterator>#include<algorithm>#include<string>#include<vector>#include<set>#include<map>#include<deque>#include<queue>#include<stack>#include<list>typedef long long ll;typedef unsigned long long llu;const int maxn = 100000 + 10;const int inf = 0x3f3f3f3f;const double pi = acos(-1.0);const double eps = 1e-8;using namespace std;int vis[10];int a;int main(){    int n;    while(scanf("%d",&n) == 1){        memset(vis,0,sizeof vis);        for (int i = 0; i < n; ++i)        {            scanf("%d",&a);            vis[a]++;        }        if (vis[5] || vis[7]){            printf("-1\n");            continue;        }        if (vis[1] != n/3)printf("-1\n");        else {            if (vis[2] - vis[4] < 0 || vis[2] - vis[4] != vis[6] - vis[3] || vis[6] < vis[3])printf("-1\n");            else {                for (int i = 0; i < vis[4]; ++i)printf("1 2 4\n");                for (int i = 0; i < vis[6]-vis[3]; ++i)printf("1 2 6\n");                for (int i = 0; i < vis[3]; ++i)printf("1 3 6\n");            }        }    }    return 0;}