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>
#define REP(i,n) for(int i=0; i<(int)(n); ++i)

using namespace std;

int main(){
    int N;
    while(cin >> N && N){
        bool ok = true;
        vector<string> words(N);
        REP(i, N) cin >> words[i];
        bool g[30][30] = {};
        for(int i = 0; i < N; i++){
            for(int j = i + 1; j < N; j++){
                for(int k = 0; k < words[i].size(); k++){
                    if(k == words[j].size()){
                        ok = false;
                        break;
                    }
                    if(words[i][k] != words[j][k]){
                        g[ words[i][k] - 'a' ][ words[j][k] - '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]) ok = false;
        cout << (ok ? "yes" : "no") << endl;
    }
    return 0;
}