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 for_(i,a,b) for(int i=(a);i<(b);++i)

int n;
string str[555];

map< char, set< char > > arc;
vector< bool > vis;

bool dfs(char c) {
	//cerr << c << endl;
	if (vis[c - 'a']) return true;
	vis[c - 'a'] = true;
	
	for (char nc : arc[c]) if (dfs(nc)) return true;
	
	vis[c - 'a'] = false;
	
	return false;
}

void solve() {
	for_(i,1,n) {
		for_(j,0,i) {
			if (str[j].size() <= str[i].size()) continue;
			
			if (str[j].substr(0, str[i].size()) == str[i]) {
				cout << "no" << endl;
				return;
			}
		}
	}
	
	for_(k,0,10) {
		int p = -1;
		
		for_(i,0,n) {
			if ((int)str[i].size() < k + 1) continue;
			
			if (p != -1) {
				if (str[p].substr(0, k) == str[i].substr(0, k)) {
					if (str[p][k] != str[i][k]) {
						arc[str[i][k]].insert(str[p][k]);
					}
				}
			}
			
			p = i;
		}
	}
	
	for_(i,0,26) {
		//cerr << "-----" << endl;
		if (dfs('a' + i)) {
			cout << "no" << endl;
			return;
		}
	}
	
	cout << "yes" << endl;
}

int main() {
	while (cin >> n, n) {
		for_(i,0,n) cin >> str[i];
		arc.clear();
		vis.assign(26, false);
		solve();
	}
}