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;

using Graph=vector<vector<int>>;
const int SIZE='z'-'a'+1;
bool solve(int n){
    vector<string> s(n);
    for(int i=0;i<n;i++) cin>>s[i];
    Graph g(SIZE);
    for(int i=0;i+1<n;i++){
        for(int j=0;j<=min(s[i].size(),s[i+1].size());j++){

            if(j==min(s[i].size(),s[i+1].size())){
                if(s[i].size()<=s[i+1].size()){
                    break;
                }
                else{
                    return false;
                }
            }
            if(s[i][j]!=s[i+1][j]){
                g[s[i][j]-'a'].push_back(s[i+1][j]-'a');
                break;
            }
        }
    }

    queue<int> que;
    vector<int> d(SIZE);
    for(int i=0;i<SIZE;i++) for(int j=0;j<g[i].size();j++) d[g[i][j]]++;
    for(int i=0;i<SIZE;i++) if(d[i]==0) que.push(i);
    int cnt=0;
    while(!que.empty()){
        int v=que.front(); que.pop();
        cnt++;
        for(int i=0;i<g[v].size();i++){
            d[g[v][i]]--;
            if(d[g[v][i]]==0){
                que.push(g[v][i]);
            }
        }
    }
    return cnt==SIZE;
}
int main(){
    int n;
    while(cin>>n,n){
        cout<<(solve(n) ? "yes" : "no")<<endl;
    }
    return 0;
}
