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
#define _USE_MATH_DEFINES

#include <cstdio>
#include <cstdlib>
#include <string.h>
#include <iostream>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>

using namespace std;

typedef pair<long long int, long long int> P;
long long int INF = 1e18;

#define MAX_V 11000

int V;
vector<int> G[MAX_V]; // グラフの隣接リスト表現
vector<int> rG[MAX_V];// 辺の向きを逆にしたグラフ
vector<int> vs;       // 帰りがけ順の並び
bool used[MAX_V];     // すでに調べたか
int cmp[MAX_V];       // 属する強連結成分のトポロジカル順序

// from から to への辺を張る関数
void add_edge(int from, int to){
	G[from].push_back(to);
	rG[to].push_back(from);
}

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);
		}
	}
}

// 強連結成分分解を行う関数、返り値は強連結成分の個数
int scc(){
	memset(used, 0, sizeof(used));
	vs.clear();
	for(int v = 0; v < V; v++){
		if(!used[v]){
			dfs(v);
		}
	}
	memset(used, 0, sizeof(used));
	int k = 0;
	for(int i = vs.size() - 1; i >= 0; i--){
		if(!used[vs[i]]){
			rdfs(vs[i], k++);
		}
	}
	return k;
}

int main(){
	
	while(true){
		
		int n;
		cin >> n;
		
		if(n == 0){
			break;
		}
		
		V = 26;
		for(int i = 0; i < MAX_V; i++){
			G[i].clear();
			rG[i].clear();
		}
		
		string str, str_pre;
		cin >> str_pre;
		
		bool flag = true;
		
		for(int i = 0; i < n - 1; i++){
			cin >> str;
			for(int j = 0; j <= min(str.size(), str_pre.size()); j++){
				if(j == min(str.size(), str_pre.size())){
					if(str.size() < str_pre.size()){
						flag = false;
					}
					break;
				}
				if(str[j] != str_pre[j]){
					add_edge(str_pre[j] - 'a', str[j] - 'a');
					break;
				}
			}
			str_pre = str;
		}
		
		flag &= (scc() == 26);
		
		if(flag){
			cout << "yes" << endl;
		}else{
			cout << "no" << endl;
		}
	}
	return 0;
}
