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;

#define int long long
typedef vector<int>vint;
typedef pair<int,int>pint;
typedef vector<pint>vpint;
#define rep(i,n) for(int i=0;i<(n);i++)
#define reps(i,f,n) for(int i=(f);i<(n);i++)
#define all(v) (v).begin(),(v).end()
#define each(it,v) for(__typeof((v).begin()) it=(v).begin();it!=(v).end();it++)
#define pb push_back
#define fi first
#define se second
template<typename A,typename B>inline void chmin(A &a,B b){if(a>b)a=b;}
template<typename A,typename B>inline void chmax(A &a,B b){if(a<b)a=b;}

int N;
string S[500];

vint G[26];
void solve(){
    rep(i,26)G[i].clear();
    rep(i,N)reps(j,i+1,N){
        int k=0;
        while(k<S[i].size()&&k<S[j].size()&&S[i][k]==S[j][k])k++;
        if(k==S[i].size()||k==S[j].size()){
            if(S[i].size()>S[j].size()){
                cout<<"no"<<endl;
                return;
            }
            continue;
        }
        G[S[i][k]-'a'].pb(S[j][k]-'a');
    }
    rep(i,26){
        sort(all(G[i]));
        G[i].erase(unique(all(G[i])),G[i].end());
    }

    vint deg(26);
    rep(i,26)rep(j,G[i].size())deg[G[i][j]]++;
    queue<int>que;
    rep(i,26)if(deg[i]==0)que.push(i);
    while(que.size()){
        int v=que.front();
        que.pop();
        for(auto u:G[v]){
            if(--deg[u]==0)que.push(u);
        }
    }
    rep(i,26)if(deg[i]){
    cout<<"no"<<endl;return;}
    cout<<"yes"<<endl;
}

signed main(){
    while(cin>>N,N){
        rep(i,N)cin>>S[i];
        solve();
    }
    return 0;
}