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<algorithm>
using namespace std;
 
#define INF 100000000
bool p[26][26];
int d[26];
int n;
string s[510];
 
int main(){
  while(cin >> n,n){
    for(int i=0;i<n;i++)cin >> s[i];
 
    for(int i=0;i<26;i++)
      for(int j=0;j<26;j++)p[i][j] = false;
 
    bool f = true;
    for(int i=0;i<n;i++){
      for(int j=i+1;j<n;j++){
	int lim = min(s[i].size(),s[j].size());
	int k;
	for(k=0;k<lim;k++){
	  if(s[i][k] != s[j][k]){
	    p[s[i][k]-'a'][s[j][k]-'a'] = true;
	    break;
	  }
	}
	if(k==lim && s[i].size()>s[j].size()){
	  f = false;
	  break;
	}
      }
    }
 
    for(int k=0;k<26;k++){
      for(int i=0;i<26;i++){
	for(int j=0;j<26;j++){
	  if(p[i][k] && p[k][j])p[i][j] = true;
	}
      }
    }
  
    for(int i=0;i<26;i++)if(p[i][i])f = false;
 
    if(!f)cout << "no\n";
    else cout << "yes\n";
  }
}