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;

int N;
string S[555];
vector<int> G[30];
vector<int> rG[30];
bool used[30];

void dfs(int v,vector<int>& vs){
  used[v] = true;
  for(int i=0;i<(int)G[v].size();i++){
    if( used[G[v][i]] ) continue;
    dfs( G[v][i] , vs );
  }
  vs.push_back( v );
  return ;
}
void rdfs(int v,int t){
  used[v] = true;
  for(int i=0;i<(int)rG[v].size();i++){
    if( used[rG[v][i]] ) continue;
    rdfs( rG[v][i], t );
  }  
}

int scc(){
  memset(used,0,sizeof(used));
  vector<int> vs;
  for(int i=0;i<26;i++)
    if( !used[i] ) dfs(i,vs);
  memset(used,0,sizeof(used));
  int t=0;
  for(int i=25;i>-1;i--)
    if( !used[vs[i]] ) rdfs( vs[i], t++ );
  return t;
}

void init(){
  for(int i=0;i<26;i++){
    G[i].clear(); rG[i].clear();
  }    
}

void add_edge(char u, char v){
  if( u == '#' || v == '#' ) return;
  G[u-'a'].push_back( v-'a' );
  rG[v-'a'].push_back( u-'a' );
}

void view(){
  for(int i=0;i<26;i++){
    char c = i+'a';
    cout << c << ": " << endl;
    for(int j=0;j<(int)G[i].size();j++){
      cout << " -> " << (char)(G[i][j]+'a') << endl;
    }
  }
}

bool makeg(int l,int h,int f){
  // cout << l<<" " << h << " " << f<< endl;
  if( l>=h ) return true;
  char c=S[l][f];
  int pd = l;
  if( c == '#' )
    return makeg(l+1,h,f);  
  for(int i=l+1;i<=h;i++){
    //cout << S[i] << endl;
    if( S[i][f] == '#' ){
      if( c != '#' ) return false;
      return makeg(i+1,h,f);
    }
    //    cout << c << "  "<< S[i][f] << endl;
    if( S[i][f] != c ){
      if( !makeg(pd,i-1,f+1) ) return false;
      add_edge( S[i][f], c );
      c = S[i][f];
      pd = i;
    }
  }
  return makeg(pd,h,f+1);
}

bool make_graph(){
  return makeg(0,N-1,0);
  view();
}

int main(){
  while( cin >> N && N ){
    for(int i=0;i<N;i++){
      cin >> S[i];
      S[i]+="#";
    }
    init();
    if( !make_graph() || scc() < 26 ) cout << "no" << endl;
    else cout << "yes" << endl;
  }
}