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
/*
4
abd
abcd
c
b
0
=>yes, where a=1,b=4,c=3,d=2
*/

#include <vector>
#include <map>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
int main(){
	int n,i,j;
	string s;
	for(;cin>>n,n;){
		map<pair<char,char>,int>m;
		vector<string>v;
		for(;n;n--){
			cin>>s;
			for(i=0;i<v.size();i++){
				for(j=0;j<min(v[i].length(),s.length());j++){
					if(v[i][j]==s[j])continue;
					if(m[make_pair(v[i][j],s[j])])goto fail;
					m[make_pair(s[j],v[i][j])]=1;break;
				}
				if(j==s.length()&&v[i].length()>s.length())goto fail;
			}
			v.push_back(s);
		}
		cout<<"yes"<<endl;continue;
		fail:
		for(n--;n;n--)cin>>s;
		cout<<"no"<<endl;
	}
}