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<algorithm>
#include<vector>
#include<string>
using namespace std;

vector<int> edge[150];
bool come[150];
int dfs(int x, int k, int p = 1){
	if(k == x && p != 0)return -1;
	if(come[x])return 0;
	come[x] = true;
	for(int i = 0;i < edge[x].size();i++){
		if(dfs(edge[x][i], k) == -1)return -1;
	}
	return 0;
}

bool ok(){
	for(int i = 'a';i <= 'z';i++){
		fill(come, come + 150, false);
		if(dfs(i, i, 0) == -1)return false;
	}
	return true;
}

int main(){
	int n;
	string str, last;
	while(cin >> n, n){
	
		bool fail = false;
		for(int i = 'a';i <= 'z';i++)edge[i].clear();
		for(int i = 0;i < n;i++){
			cin >> str;
			if(i){
				for(int i = 0;i < last.size();i++){
					if(i == str.size()){
						fail = true;
						break;
					}
					if(str[i] != last[i]){
						edge[last[i]].push_back(str[i]);
						break;
					}
				}
			}
			last = str;
		}
		
		cout << ((!fail&&ok())?"yes":"no") << endl;
	}
	return 0;
}