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<cstring>
#include<string>
#include<cstdio>
using namespace std;
int n;
string S[505];
int G[30][30];
int main()
{
    //freopen("data.in","r",stdin);
    while(cin>>n)
    {
        if(n==0) break;

        for(int i=1;i<=n;i++) cin>>S[i];

        memset(G,0,sizeof(G));
        bool ok=true;
        for(int i=1;i<=n;i++)
            for(int j=i+1;j<=n;j++)
        {
            bool t=true;

            for(int k=0;(k<S[i].size())&&(k<S[j].size());k++)
            {
                if(S[i][k]!=S[j][k])
                {
                    t=false;
                    G[S[j][k]-'a'][S[i][k]-'a']=1;
                    break;
                }
            }
            if(t&&(S[i].size()>S[j].size())) ok=false;
        }

        for(int i=0;i<26;i++)
            for(int j=0;j<26;j++)
                if(i!=j)
            {
                if(G[i][j]==1&&G[j][i]==1) ok=false;
            }
        if(ok) cout<<"yes"<<endl;
        else cout<<"no"<<endl;

    }
    return 0;
}