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 <vector>
#include <string>

using namespace std;

#define loop(i,a,b) for(int i=(a); i<(int)(b); i++)
#define rep(i,b) loop(i,0,b)

bool isPrefix(const string& a, const string &b){
    if(a.size() <= b.size()){
        rep(i, a.size()){
            if(a[i] == b[i]) continue;
            else return false;
        }
    } else {
        return false;
    }
    return true;
}

bool solve(int n){
    vector<string> v(n);
    int G[26][26] = {};
    rep(i, n) cin >> v[i];
    loop(i, 1, n){
        if(isPrefix(v[i - 1], v[i])){
            continue;
        } else if(isPrefix(v[i], v[i - 1])){
            return false;
        } else {
            rep(j, min(v[i].size(), v[i - 1].size())){
                char a = v[i][j], b = v[i - 1][j];
                if(a == b){
                    continue;
                } else {
                    G[a - 'a'][b - 'a'] = true;
                    break;
                }
            }
        }
    }
    rep(k, 26) rep(i, 26) rep(j, 26){
        G[i][j] |= G[i][k] && G[k][j];
    }
    
    rep(i, 26){
        if(G[i][i])	return false;
    }
    return true;
}

int main(){
    int n;
    while(cin >> n, n){
        if(solve(n)){
            cout << "yes";
        } else {
            cout << "no";
        }
        cout << endl;
    }
}