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;

#define rep(i, n) for(int i = 0; i < (n); i++)
#define repp(i, l, r) for(int i = (l); i < (r); i++)
#define per(i, n) for(int i = ((n)-1); i >= 0; i--)
#define perr(i, l, r) for(int i = ((r)-1); i >= (l); i--)
#define all(x) (x).begin(),(x).end()
#define MOD 1000000007
#define IINF 1000000000
#define LINF 1000000000000000000
#define SP <<" "<<
#define CYES cout<<"Yes"<<endl
#define CNO cout<<"No"<<endl
#define CFS cin.tie(0);ios::sync_with_stdio(false)

typedef long long LL;
typedef long double LD;

int main(){
  while(1){
    int n;
    cin >> n;
    if(n==0) return 0;
    vector<vector<int>> e(26);
    vector<int> deg(26,0);
    vector<string> s(n);
    queue<int> q;
    rep(i,n){
      cin >> s[i];
    }
    rep(i,n){
      rep(j,i){
        rep(k,min(s[i].length(),s[j].length())){
          if(s[i][k]!=s[j][k]){
            e[s[j][k]-'a'].push_back(s[i][k]-'a');
            deg[s[i][k]-'a']++;
            goto next;
          }
        }
        if(s[i].length()<s[j].length()){
          cout << "no" << endl;
          goto nextt;
        }
        next:;
      }
    }
    rep(i,26) if(deg[i]==0) q.push(i);
    while(!q.empty()){
      int now = q.front();
      q.pop();
      for(auto nx:e[now]){
        deg[nx]--;
        if(deg[nx]==0) q.push(nx);
      }
    }
    rep(i,26){
      if(deg[i]!=0){
        cout << "no" << endl;
        goto nextt;
      }
    }
    cout << "yes" << endl;
    nextt:;
  }
  return 0;
}

