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<vector>
using namespace std;

int main(){
    int n;
    while(cin >> n, n){
        int dp[26][26] = {};
        string b;
        cin >> b;
        bool ng = false;
        for(int i = 1; i < n; i++){
            string s;   cin >> s;
            int j;
            for(j = 0; j < min(b.length(), s.length()); j++){
                if(b[j] != s[j])    break;
            }
            if(j < min(b.length(), s.length())){
                dp[b[j]-'a'][s[j]-'a'] = -1;
            }else if(j == s.length() && j < b.length()){
                ng = true;
            }
            b = s;
        }
        for(int k = 0; k < 26; k++){
            for(int i = 0; i < 26; i++){
                for(int j = 0; j < 26; j++){
                    if(dp[i][k] == 0 || dp[k][j] == 0)  continue;
                    dp[i][j] = min(dp[i][j], dp[i][k]+dp[k][j]);
                }
            }
        }
        for(int i = 0; i < 26; i++){
            ng |= (dp[i][i] < 0);
        }
        cout << (ng ? "no" : "yes") << endl;
    }
    return 0;
}
