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>
using namespace std;
using ll = long long;
#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define fi first
#define se second

int vis[30];
vector<int> G[30];
bool valid;

void dfs(int now)
{
    if(vis[now]==2)
    {
        valid=false;
        return;
    }
    else if(vis[now]==0)
    {
        vis[now]=2;
        rep(i,G[now].size()) dfs(G[now][i]);

        vis[now]=1;
    }
}

int main()
{
    int n;
    while(cin >>n,n)
    {
        vector<string> s(n);
        rep(i,n) cin >>s[i];

        rep(i,30)
        {
            vis[i]=0;
            G[i].clear();
            valid=true;
        }

        rep(i,n)for(int j=i+1; j<n; ++j)
        {
            int I=s[i].size(), J=s[j].size();
            int S=min(I,J);

            int idx=0;
            while(idx<S)
            {
                if(s[i][idx] != s[j][idx]) break;
                ++idx;
            }

            if(idx<S)
            {
                int from = s[i][idx]-'a', to = s[j][idx]-'a';
                G[from].pb(to);
            }
            else
            {
                if(I>J) valid=false;
            }
        }

        rep(i,30) dfs(i);

        cout << (valid?"yes":"no") << endl;
    }
    return 0;
}