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() {
  for(int n; cin >> n && n; ) {
    vector<string> v(n);
    for(int i = 0; i < n; ++i) cin >> v[i];
    try {
      vector<int> s, t;
      for(int i = 0; i+1 < n; ++i) {
        const string &a = v[i], &b = v[i+1];
        int l = min(a.size(), b.size());
        if(a.substr(0, l) == b.substr(0, l) &&
           a.size() > b.size()) {
          throw "no";
        }
        for(int j = 0; j < l; ++j) {
          if(a[j] != b[j]) {
            s.push_back(a[j]-'a');
            t.push_back(b[j]-'a');
            break;
          }
        }
      }
      vector<int> w(26, 0);
      for(int k = 0; k < 26; ++k) {
        bool update = false;
        for(int i = 0; i < s.size(); ++i) {
          if(w[t[i]] > w[s[i]] - 1) {
            w[t[i]] = w[s[i]] - 1;
            update = true;
          }
        }
        if(!update) throw "yes";
        if(k+1 == 26) throw "no";
      } 
    } catch(const char *s) {
      cout << s << endl;
    }
  }
  return 0;
}