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<stdio.h>
#include<algorithm>
using namespace std;
char str[510][20];
int g[30][30];
int main(){
	int a;
	while(scanf("%d",&a),a){
		for(int i=0;i<a;i++)scanf("%s",str[i]);
		for(int i=0;i<26;i++)for(int j=0;j<26;j++)g[i][j]=0;
		for(int i=0;i<26;i++)g[i][i]=1;
		bool ok=true;
		for(int i=0;i<a;i++){
			for(int j=i+1;j<a;j++){
				int k;
				for(k=0;str[i][k]&&str[j][k];k++){
					if(str[i][k]!=str[j][k]){
						g[str[i][k]-'a'][str[j][k]-'a']=1;break;
					}
				}
				if(str[i][k]&&!str[j][k])ok=false;
			}
		}
		for(int k=0;k<26;k++)for(int i=0;i<26;i++)for(int j=0;j<26;j++)
			g[i][j]|=(g[i][k]&g[k][j]);
		
		for(int i=0;i<26;i++)for(int j=0;j<26;j++)if(i!=j&&g[i][j]&&g[j][i])ok=false;
		if(ok)printf("yes\n");
		else printf("no\n");
	}
}