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>
#include<string>

using namespace std;

int main(){
  for(int n;cin>>n,n;){
    string str[500];
    int d[128][128];
    int INF=1<<29;
    fill(d[0],d[128],INF);
    for(int i=0;i<128;i++){
      d[i][i]=0;
    }
    bool f=false;
    for(int i=0;i<n;i++){
      cin>>str[i];
      for(int j=0;j<i;j++){
	int mm=min(str[i].size(),str[j].size());
	if(str[i].compare(0,mm,str[j],0,mm)==0){
	  f|=str[i].size()<str[j].size();
	}else{
	  auto x=mismatch(begin(str[i]),end(str[i]),begin(str[j]));
	  d[*x.first][*x.second]=-1;
	}
      }
    }
    for(int i=0;i<128;i++){
      for(int j=0;j<128;j++){
	for(int k=0;k<128;k++){
	  if(d[j][i]!=INF&&d[i][k]!=INF){
	    d[j][k]=min(d[j][k],d[j][i]+d[i][k]);
	  }
	}
      }
    }
    for(int i=0;i<128;i++){
      f|=d[i][i]<0;
    }
    cout<<(f?"no":"yes")<<endl;
  }
}

	   