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;

int main()
{
  int N;
  while(cin >> N, N) {
    vector< string > s(N);
    
    bool flag = true;
    for(int i = 0; i < N; i++) cin >> s[i];
    bool graph[26][26] = {{}};
    for(int i = 1; i < N; i++) {
      bool update = false;
      for(int j = 0; j < min(s[i - 1].size(), s[i].size()); j++) {
        if(s[i - 1][j] != s[i][j]) {
          graph[s[i - 1][j] - 'a'][s[i][j] - 'a'] = true;
          update = true;
          break;
        }
      }
      flag &= update | (s[i - 1].size() <= s[i].size());
    }
    for(int k = 0; k < 26; k++) {
      for(int i = 0; i < 26; i++) {
        for(int j = 0; j < 26; j++) {
          graph[i][j] |= graph[i][k] & graph[k][j];
        }
        flag &= !graph[i][i];
      }
    }
    if(flag) cout << "yes" << endl;
    else cout << "no" << endl;
  }
}