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 <sstream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <ctime>
#include <cctype>
#include <cstring>
#include <string>
#include <vector>
#include <list>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <set>
#include <bitset>
#include <numeric>
#include <complex>
#include <utility>
#include <iomanip>
#include <algorithm>
#include <functional>
using namespace std;

typedef vector<int> vint;
typedef vector<vint> vvint;
typedef vector<string> vst;
typedef pair<int,int> pint;
typedef long long ll;

#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define all(s) (s).begin(),(s).end()
#define each(i,s) for (typeof((s).begin()) i = (s).begin(); i != (s).end(); ++i)
#define debug(x) cout<<#x<<" = "<<(x)<<" (L"<<__LINE__<<")"<<endl
template<class T1, class T2> ostream& operator<<(ostream &s, pair<T1,T2> P){return s<<'<'<<P.first<<", "<<P.second<<'>';}
template<class T> ostream& operator<<(ostream &s, vector<T> P) {s<<"{ ";for(int i=0;i<P.size();++i){if(i>0)s<<", ";s<<P[i];}return s<<" }"<<endl;}
template<class T1, class T2> ostream& operator<<(ostream &s, map<T1,T2> P) {s<<"{ ";for(typeof(P.begin()) it=P.begin();it!=P.end();++it){if(it!=P.begin())s<<", ";s<<'<'<<it->first<<"->"<<it->second<<'>';}return s<<" }"<<endl;}
template<class T> void print(T P[], int n) {cout<<"{ ";for(int i=0;i<n;++i){if(i>0)cout<<", ";cout<<P[i];}cout<<" }"<<'\n';}
template<class T> void print(T P[][30], int n, int m) {for(int i=0;i<n;++i){cout<<i<<" : ";print(P[i],m);}}



int n;
vst str;

bool d[30][30];

string solve() {
    map<char,int> ma;
    for (int i = 0; i < 26; ++i) ma[(char)('a'+i)] = i;

    memset(d, 0, sizeof(d));
    for (int i = 1; i < str.size(); ++i) {
        string s1 = str[i-1];
        string s2 = str[i];
        
        for (int j = 0; j < min((int)s1.size(), (int)s2.size()); ++j) {
            if (s1[j] != s2[j]) {
                d[ma[s1[j]]][ma[s2[j]]] = true;
                break;
            }
            else if (j == min((int)s1.size(), (int)s2.size())-1 && s1.size() > s2.size()) {
                return "no";
            }
        }
    }
    
    bool ok = true;
    for (int k = 0; k < 26; ++k) {
        for (int i = 0; i < 26; ++i) {
            for (int j = 0; j < 26; ++j) {
                d[i][j] |= d[i][k] & d[k][j];
            }
        }
    }
    
    for (int i = 0; i < 26; ++i) {
        if (d[i][i]) ok = false;
    }
    
    //print(d, 26, 26);
    
    return ok ? "yes" : "no";
}

int main() {
    //freopen( "/Users/macuser/Documents/Programming/Contest/JAG Practice Asia 2012 A - Dictionary.in", "r", stdin );
    
    while (cin >> n) {
        if (n == 0) break;
        str.clear();
        for (int i = 0; i < n; ++i) {
            string temp;
            cin >> temp;
            str.pb(temp);
        }
        cout << solve() << endl;
    }
    
    return 0;
}
