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>

#define REP(i,n) for(int i=0;i<(int)(n);i++)

using namespace std;

int main() {
  while(1){
    int n;
    cin>>n;
    if(!n)break;
    vector<string> d(n);
    REP(i,n){
      string w;
      cin>>w;
      d[i] = w + string(11-w.size(), '`');
    }
    d.erase(unique(begin(d),end(d)),end(d));
    vector<vector<int>> dst(27,vector<int>(27,0));
    REP(i,n-1){
      REP(j,11){
        if(d[i][j] != d[i+1][j]) {
          dst[d[i][j]-'`'][d[i+1][j]-'`'] = 1;
          break;
        }
      }
    }
    REP(i,27)
      if(i) dst[0][i] = 1;
    REP(k,27)REP(i,27)REP(j,27) dst[i][j] = max(dst[i][j], dst[i][k] * dst[k][j]);
    bool ok = true;
    REP(i,27)
      if(dst[i][i]) ok = false;
    if (ok)
      cout << "yes" << endl;
    else
      cout << "no" << endl;
  }
  return 0;
}