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){
		bool no = false;
		int dag[128][128]={};
		string p;
		for(int i = 0 ; i < n ; i++){
			string s;
			cin >> s;
			if(i){
				int j = 0;
				while( j < min(p.size(),s.size()) && s[j] == p[j] ) j++;
				
				if( j != p.size() && j != s.size() ){
					dag[p[j]][s[j]] = true;
				}else{
					if( p.size() > s.size() )
						no = true;
				}
			}
			p = s;
		}
		if( no ){
			cout << "no" << endl;
			continue;
		}
		
		for(int i = 'a' ; i <= 'z' ; i++){
			for(int j = 'a' ; j <= 'z' ; j++){
				for(int k = 'a' ; k <= 'z' ; k++){
					dag[j][k] |= dag[j][i] & dag[i][k];
				}
			}
		}
		string yes = "yes";
		for(int i = 'a' ; i <= 'z' ; i++)
			if( dag[i][i] ) yes = "no";
		cout << yes << endl;
	}
}