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


vector<int> G[26];
vector<int> rG[26];
vector<int> vs;
bool used[26];
int cmp[26];

void init(){
	for(int i = 0; i < 26; ++i){
		G[i].clear();
		rG[i].clear();
		vs.clear();
		used[i] = false;
		cmp[i] = 0;
	}
}


void add_edge(int f, int t){
	G[f].push_back(t);
	rG[t].push_back(f);
}

void dfs(int v){
	used[v] = true;
	for(int i = 0; i < G[v].size(); ++i){
		if( !used[G[v][i]] ) dfs(G[v][i]);
	}
	vs.push_back(v);
}

void rdfs(int v, int k){
	used[v] = true;
	cmp[v] = k;
	for(int i = 0; i < rG[v].size(); ++i){
		if( !used[rG[v][i]] ) rdfs(rG[v][i], k);
	}
}

int scc(){
	for(int i = 0; i < 26; ++i) used[i] = false;
	vs.clear();
	for(int v = 0; v < 26; ++v){
		if( !used[v] ) dfs(v);
	}
	for(int i = 0; i < 26; ++i) used[i] = false;
	int k = 0;
	for(int i = (int)vs.size() - 1; i >= 0; --i){
		if( !used[vs[i]] ) rdfs(vs[i], k++);
	}
	return k;
}


bool solve(const vector<string> &v){
	init();
	int n = v.size();
	
	for(int i = 0; i < n - 1; ++i){
		bool f = true;
		for(int j = 0; j < v[i].size(); ++j){
			if( j >= v[i+1].size() ){
				return false;
			}
			if( v[i][j] != v[i+1][j] ){
				add_edge(v[i][j] - 'a', v[i+1][j] - 'a');
				break;
			}
		}
	}

	int k = scc();
	return k == 26;
}


int main(){
	int n;
	while( cin >> n, n != 0 ){
		vector<string> v(n);
		for(int i = 0; i < n; ++i){
			cin >> v[i];
		}

		cout << (solve(v) ? "yes\n" : "no\n");
	}
}