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 <string>
#include <vector>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
int n; vector<string> s;
bool ok(int d, int l, int r) {
	if (s[l].size() == d) l++;
	for (int i = l; i < r; i++) {
		if (s[i].size() == d) return false;
	}
	bool ret = true;
	for (int p = l, i = l + 1; i <= r; i++) {
		if (i == r || s[i - 1][d] != s[i][d]) {
			if (!ok(d + 1, p, i)) ret = false;
			p = i;
		}
	}
	return ret;
}
bool cycle_detection(vector<vector<int> > &G) {
	vector<bool> vis(G.size());
	function<bool(int, int, bool)> rec = [&](int pos, int start, bool flag) {
		if (pos == start && flag) return true;
		if (flag) vis[pos] = true;
		bool ret = false;
		for (int e : G[pos]) {
			if (!vis[e]) ret = ret || rec(e, start, true);
		}
		return ret;
	};
	for (int i = 0; i < G.size(); i++) {
		fill(vis.begin(), vis.end(), false);
		if (rec(i, i, false)) return true;
	}
	return false;
}
int main() {
	while (cin >> n, n) {
		s = vector<string>(n);
		for (int i = 0; i < n; i++) cin >> s[i];
		s.erase(unique(s.begin(), s.end()), s.end()); n = s.size();
		bool res = ok(0, 0, n);
		if (!res) cout << "no" << endl;
		else {
			vector<vector<int> > G(26);
			for (int i = 1; i < n; i++) {
				for (int j = 0; j < min(s[i - 1].size(), s[i].size()); j++) {
					if (s[i - 1][j] != s[i][j]) {
						G[s[i - 1][j] - 97].push_back(s[i][j] - 97);
						break;
					}
				}
			}
			cout << (cycle_detection(G) ? "no" : "yes") << endl;
		}
	}
	return 0;
}