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<vector>
#include<algorithm>
using namespace std;
int n;
string ss[500];
vector<int>G[26];
bool used[26];
void dfs(int u)
{
	for(int a:G[u])
	{
		if(!used[a])
		{
			used[a]=1;
			dfs(a);
		}
	}
}
int main()
{
	while(cin>>n,n)
	{
		for(int i=0;i<n;i++)cin>>ss[i];
		bool flag=1;
		for(int i=0;i<26;i++)G[i].clear();
		for(int i=0;i<n-1;i++)
		{
			string s=ss[i],t=ss[i+1];
			int id=0;
			while(id<s.size()&&id<t.size()&&s[id]==t[id])id++;
			if(id>=s.size());
			else if(id>=t.size())
			{
				flag=0;
			}
			else
			{
				G[s[id]-'a'].push_back(t[id]-'a');
			}
		}
		if(flag)
		{
			for(int i=0;i<26;i++)
			{
				G[i].erase(unique(G[i].begin(),G[i].end()),G[i].end());
				for(int j=0;j<26;j++)used[j]=0;
				dfs(i);
				flag&=!used[i];
			}
		}
		cout<<(flag?"yes":"no")<<endl;
	}
}

