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 <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <vector>
#include <iostream>
using namespace std;
string s[1011];
vector<int> V[33];
int main()
{
    int n;
    while(1)
    {
        for(int i = 0; i <= 26; i++)
            V[i].clear();
        scanf("%d", &n);
        if(n == 0)
            break;
        for(int i = 0; i < n; i++)
            cin>>s[i];
        int cnt = 0;
        int in[33] = {0};
        for(int i = 0; i < n; i++)
            for(int j = i + 1; j < n; j++)
            {
                int l1 = s[i].size();
                int l2 = s[j].size();
                int flag = 0;
                for(int k = 0; k < max(l1, l2); k++)
                {
                    if(k >= l1 && k <= l2)
                    {
                        break;
                    }
                    if(k >= l2 && k < l1)
                    {
                        puts("no");
                        goto loop;
                    }
                    if(s[i][k] != s[j][k])
                    {
                        V[s[i][k] - 'a'].push_back(s[j][k] - 'a');
                        in[s[j][k] - 'a']++;
                        break;
                    }
                }
            }
        while(1)
        {
            int nxt = -1;
            for(int i = 0; i < 26; i++)
                if(in[i] == 0)
                {
                    nxt = i;
                    break;
                }
            if(nxt == -1)
                break;
            in[nxt]--;
            cnt++;
            for(int i = 0; i < V[nxt].size(); i++)
                in[V[nxt][i]]--;
        }
        if(cnt == 26)
            puts("yes");
        else puts("no");
        loop:;
    }
}