Problem Statement

We found a dictionary of the Ancient Civilization Mayo (ACM) during excavation of the ruins. After analysis of the dictionary, we revealed they used a language that had not more than 26 letters. So one of us mapped each letter to a different English alphabet and typed all the words in the dictionary into a computer.

How the words are ordered in the dictionary, especially whether they are ordered lexicographically, is an interesting topic to many people. As a good programmer, you are requested to write a program to judge whether we can consider the words to be sorted in a lexicographical order.

Note: In a lexicographical order, a word always precedes other words it is a prefix of. For example, `ab` precedes `abc`, `abde`, and so on.



Input

The input consists of multiple datasets. Each dataset is formatted as follows:


n
string_1
...
string_n


Each dataset consists of n+1 lines. The first line of each dataset contains an integer that indicates n (1 \leq n \leq 500). The i-th line of the following n lines contains string_i, which consists of up to 10 English lowercase letters.

The end of the input is `0`, and this should not be processed.

Output

Print either `yes` or `no` in a line for each dataset, in the order of the input. If all words in the dataset can be considered to be ordered lexicographically, print `yes`. Otherwise, print `no`.

Example

Input

4
cba
cab
b
a
3
bca
ab
a
5
abc
acb
b
c
c
5
abc
acb
c
b
b
0


Output

yes
no
yes
no
#include <bits/stdc++.h>

using namespace std;

bool solve(int n)
{
  string str[505];
  for(int i = 0; i < n; i++) cin >> str[i];

  bool G[33][33] = {{false}};
  for(int i = 0; i < n-1; i++) {
    bool flag = (str[i].size() <= str[i+1].size());
    for(int j = 0; j < min(str[i].size(), str[i+1].size()); j++) {
      if(str[i][j] != str[i+1][j]) {
	G[str[i][j]-'a'][str[i+1][j]-'a'] = true;
	flag = true;
	break;
      }
    }
    if(!flag) return true;
  }
  
  for(int i = 0; i < 26; i++) {
    for(int j = 0; j < 26; j++) {
      for(int k = 0; k < 26; k++) {
	G[j][k] |= G[j][i] & G[i][k];
      }
    }
  }
  
  bool ans = false;
  for(int i = 0; i < 26; i++) ans |= G[i][i];

  return ans;
}

int main()
{
  int n;
  while(cin >> n, n) {
    cout << (solve(n) ? "no" : "yes") << endl;
  }
  return 0;
}