DNA Sorting

Problem Description

One measure of ``unsortedness‘‘ in a sequence is the number of pairs of entries that are out of order with respect to each other. For instance, in the letter sequence ``DAABEC‘‘, this measure is 5, since D is greater than four letters to its right and E is greater than one letter to its right. This measure is called the number of inversions in the sequence. The sequence ``AACEDGG‘‘ has only one inversion (E and D)--it is nearly sorted--while the sequence ``ZWQM‘‘ has 6 inversions (it is as unsorted as can be--exactly the reverse of sorted).

You are responsible for cataloguing a sequence of DNA strings (sequences containing only the four letters A, C, G, and T). However, you want to catalog them, not in alphabetical order, but rather in order of ``sortedness‘‘, from ``most sorted‘‘ to ``least sorted‘‘. All the strings are of the same length.


This problem contains multiple test cases!

The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

 

Input

The first line contains two integers: a positive integer n (0 < n <= 50) giving the length of the strings; and a positive integer m (1 < m <= 100) giving the number of strings. These are followed by m lines, each containing a string of length n.

 

Output

Output the list of input strings, arranged from ``most sorted‘‘ to ``least sorted‘‘. If two or more strings are equally sorted, list them in the same order they are in the input file.

 

Sample Input

 

Sample Output

#include <iostream>#include <string>#include <algorithm>using namespace std;struct DNA{    string s;    int d;};int cmp(DNA a,DNA b){    if(a.d!=b.d)return a.d<b.d;    else return 0;}int main(){        int t;    int n,m;    DNA dna[80];    cin>>t;    while(t--)    {        cin>>n>>m;        int i,j;        for(i = 0; i < m; ++i){            cin>>dna[i].s;            dna[i].d = 0;             for(j = 1; j < n; ++j)            {                for(int k = 0; k < j ;++k)                {                    if(dna[i].s[j] < dna[i].s[k])                        dna[i].d++;                }            }        }        sort(dna,dna+m,cmp);        for(i = 0; i < m; ++i)            cout<<dna[i].s<<endl;    }    return 0;}