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<string>
#include<algorithm>
#include<vector>
#include<iomanip>
#include<math.h>
#include<complex>
#include<queue>
#include<deque>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<functional>
#include<assert.h>
#include<numeric>
using namespace std;
#define REP(i,m,n) for(int i=(int)(m) ; i < (int) (n) ; ++i )
#define rep(i,n) REP(i,0,n)
using ll = long long;
const int inf=1e9+7;
const ll longinf=1LL<<60 ;
const ll mod=1e9+7 ;


void solve(int n){
    string s[n];
    rep(i,n)cin>>s[i];
    vector<int> v[26];
    vector<int> cnt(26);
    rep(i,n-1){
        int m=min(s[i].size(),s[i+1].size());
        bool used=0;
        rep(j,m){
            if(s[i][j]!=s[i+1][j]){
                v[s[i][j]-'a'].push_back(s[i+1][j]-'a');
                cnt[s[i+1][j]-'a']++;
                used=true;
                break;
            }
        }
        if(!used&&s[i].size()>m){
            cout<<"no"<<endl;
            return;
        }
    }
    queue<int> q;
    int ans=0;
    rep(i,26){
        if(!cnt[i])++ans,q.push(i);
    }
    while(q.size()){
        int x=q.front();q.pop();
        for(auto to : v[x]){
            if(--cnt[to]==0)++ans,q.push(to);
        }
    }
    if(ans==26)cout<<"yes"<<endl;
    else cout<<"no"<<endl;
}
int main(){
    int n;
    while(cin>>n,n)solve(n);
    return 0;
}
