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.

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 (0 < 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''. Since two strings can be equally sorted, then output them according to the orginal order.

Sample Input

10 6
AACATGAAGG
TTTTGGCCAA
TTTGGCCAAA
GATCAGATTT
CCCGGGGGGA
ATCGATGCAT

Sample Output

CCCGGGGGGA
AACATGAAGG
GATCAGATTT
ATCGATGCAT
TTTTGGCCAA
TTTGGCCAAA

分析：

问题的关键在于如何求解一个序列的无序度，最差肯定是O（n2）的，这里我们可以从归纳的角度分析问题，如果已知前n-1个字母序列的无序度，那么对于第n个字母，我们需要和前n-1个字母比较来更新，问题似乎没变，还是一样的复杂度，但是这里我们只有ACGT四个字母，我们想如何用更强的归纳来减少问题的规模，对于第n个字母，我们不需要显式的和前面n-1个字母比较，只要知道前面有几个A，几个C，几个G，几个T，我们就可以直接更新无序度了，如果为A，则新的无序度需要加上前n-1个字母的CGT个数，使用更强的归纳解决问题！

#include 
using namespace std;
int find_sortedness(char *seq,int length)
{
 int numA=0,numC=0,numG=0,numT=0;
 int unsort=0;
 for(int i=0;i>length>>nums;
 char** a=new char *[nums];
 int* results=new int[nums];
 int* print=new int[nums];
 for(int i=0;i>a[i][j];
  }
  results[i]=find_sortedness(a[i],length);
  print[i]=i;
 }
 for(int i=0;ii;j--)
  {
   if(results[j]