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>
#define N 500
#define C 26
using namespace std;
int n,slen[N],edge[C][C],f[N][N];
int ans;
string s[N];

int main(){
  while(1){
    cin>>n;
    if(!n)break;
    ans=1;
    for(int i=0;i<n;i++){
      cin>>s[i];
      slen[i]=s[i].size();
    }
    for(int i=0;i<n;i++)
      for(int j=0;j<n;j++)f[i][j]=0;
    for(int i=0;i<C;i++)
      for(int j=0;j<C;j++)
	edge[i][j]=0;
    for(int i=0;i<n;i++){
      for(int j=0;j<10;j++){
	if(slen[i]<j+1)continue;
	for(int k=i+1;k<n;k++){
	  if(!f[i][k]&&s[i][j]==s[k][j])
	    if(slen[k]==j+1&&slen[i]>slen[k])
	      ans=0;
	  if(f[i][k])continue;
	  if(slen[k]<j+1)continue;
	  int i1=s[i][j]-'a';
	  int i2=s[k][j]-'a';
	  if(i1==i2)continue;
	  if(i1!=i2)f[i][k]=1;
	  edge[i1][i2]=1;
	}
      }
    }
    for(int i=0;i<C;i++)
      for(int j=0;j<C;j++)
	if(edge[i][j]&&edge[j][i])
	  ans=0;
    if(ans)cout<<"yes"<<endl;
    else cout<<"no"<<endl;
  }
  return 0;
}