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 <fstream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <algorithm>
using namespace std;
 
#define REP(i,n) for(int i=0; i<n; ++i)
#define FOR(i,a,b) for(int i=a; i<=b; ++i)
#define FORR(i,a,b) for (int i=a; i>=b; --i)
#define pi acos(-1.0)
 
typedef long long ll;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef vector<VI> VVI;
typedef pair<int,int> P;
typedef pair<ll,ll> PL;

int main(){
	int n;
	while (cin >> n && n){
		bool f = 1;
		vector<string> s(n);
		REP(i,n) cin >> s[i];
		VVI e(26,VI(26));
		REP(i,n-1){
			string a = s[i], b = s[i+1];
			REP(j,min(a.length(), b.length())){
				if (a[j] != b[j]){
					e[a[j]-'a'][b[j]-'a'] = 1;
					break;
				}
				if (j == min(a.length(), b.length())-1 && a.length() > b.length())
					f = 0;
			}
		}
		REP(i,26){
			VI visit(26);
			queue<int> que;
			que.push(i);
			while (!que.empty()){
				int x = que.front();
				que.pop();
				REP(j,26){
					if (!e[x][j] || visit[j]) continue;
					visit[j] = 1;
					que.push(j);
				}
			}
			if (visit[i]) f = 0;
		}
		cout << (f ? "yes" : "no") << endl;
	}
    return 0;
}