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;
bool f;
bool e[26][26];
void dfs(int x,int d){
  //cout << x << ":" << d << endl;
  if(d>26) {
    f=false;
    return;
  }
  if(!f) return;
  for(int i=0;i<26;i++){
    if(e[x][i]) dfs(i,d+1);
  }
}
int main(){
  int n;
  while(cin>>n,n){
    int i,j,k;
    vector<string> v;
    string s;
    memset(e,false,sizeof(e));
    for(i=0;i<n;i++){
      cin >> s;
      v.push_back(s);
    }
    f=true;
    for(i=0;i<n;i++){
      for(j=i+1;j<n;j++){
	if(v[i].find(v[j])==0&&v[i].size()>v[j].size()) f=false;
	int x=min(v[i].size(),v[j].size());
	for(k=0;k<x;k++){
	  if(v[i][k]==v[j][k]) continue;
	  e[v[i][k]-'a'][v[j][k]-'a']=true;
	  break;
	}
      }
    }
    for(i=0;i<26;i++){
      if(f)dfs(i,0);
    }
    if(f) cout << "yes" << endl;
    else cout << "no" << endl;
  }
  return 0;
}