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 <iostream>
#include <string>
#include <queue>
#include <vector>

using namespace std;

class Graph {
  typedef vector<int> Vector;
  typedef vector<Vector> Matrix;
  typedef queue<int> IntQueue;
  public:
    Graph(int size): size_(size), next_(size) {
    }
    void add_edge(int a, int b) {
      next_[a].push_back(b);
    }
    bool exists_path(int a, int b) {
      vector<bool> has_visited(size_, false);
      IntQueue q;
      q.push(a);
      has_visited[a] = true;
      while (not q.empty()) {
        int x = q.front();
        q.pop();
        for (size_t i = 0; i < next_[x].size(); ++i) {
          int y = next_[x][i];
          if (not has_visited[y]) {
            if (y == b) {
              return true;
            }
            has_visited[y] = true;
            q.push(y);
          }
        }
      }
      return false;
    }
  private:
    int size_;
    Matrix next_;
};

bool solve(int n, string s[]) {
  Graph g(26);
  for (int i = 0; i < n - 1; ++i) {
    size_t pos = 0;
    while (pos < s[i].length() and s[i][pos] == s[i + 1][pos]) {
      ++pos;
    }
    if (pos == s[i].length()) {
      continue;
    } else if (pos == s[i + 1].length()) {
      return false;
    }
    g.add_edge(s[i][pos] - 'a', s[i + 1][pos] - 'a');
    for (int i = 0; i < 26; ++i) {
      for (int j = i + 1; j < 26; ++j) {
        if (g.exists_path(i, j) and g.exists_path(j, i)) {
          return false;
        }
      }
    }
  }
  return true;
}

int main() {
  int n;
  string s[500];
  while (true) {
    cin >> n;
    if (n == 0) {
      return 0;
    }
    for (int i = 0; i < n; ++i) {
      cin >> s[i];
    }
    if (solve(n, s)) {
      cout << "yes" << endl;
    } else {
      cout << "no" << endl;
    }
  }
}