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<deque>
#include<queue>
#include<vector>
#include<algorithm>
#include<iostream>
#include<set>
#include<cmath>
#include<tuple>
#include<string>
#include<chrono>
#include<functional>
#include<iterator>
#include<random>
#include<unordered_set>
#include<array>
#include<map>
#include<iomanip>
#include<assert.h>
#include<list>
#include<bitset>
#include<stack>
#include<memory>
using namespace std;
using namespace std::chrono;
typedef long long int llint;
typedef double lldo;
#define mp make_pair
#define mt make_tuple
#define pub push_back
#define puf push_front
#define pob pop_back
#define pof pop_front
#define fir first
#define sec second
#define res resize
#define ins insert
#define era erase
/*cout<<fixed<<setprecision(20);cin.tie(0);ios::sync_with_stdio(false);*/
const llint mod=1000000007;
const llint big=2.19e18+1;
const long double pai=3.141592653589793238462643383279502884197;
const long double eps=1e-15;
template <class T,class U>bool mineq(T& a,U b){if(a>b){a=b;return true;}return false;}
template <class T,class U>bool maxeq(T& a,U b){if(a<b){a=b;return true;}return false;}
llint gcd(llint a,llint b){if(a%b==0){return b;}else return gcd(b,a%b);}
llint lcm(llint a,llint b){if(a==0){return b;}return a/gcd(a,b)*b;}
template<class T> void SO(T& ve){sort(ve.begin(),ve.end());}
template<class T> void REV(T& ve){reverse(ve.begin(),ve.end());}
template<class T>llint LBI(vector<T>&ar,T in){return lower_bound(ar.begin(),ar.end(),in)-ar.begin();}
template<class T>llint UBI(vector<T>&ar,T in){return upper_bound(ar.begin(),ar.end(),in)-ar.begin();}
bool solve(void){
	int n,i,j;cin>>n;
	if(n==0){return false;}
	string mae="#";
	bool dame=false;
	int zyun[26][26]={};
	int zi[26]={};
	for(i=0;i<n;i++){
		
		string now;cin>>now;
		now.pub('#');
		j=0;
		while(mae[j]!='#'&&mae[j]==now[j]){j++;}
		if(mae[j]=='#'){mae=now;continue;}
		if(now[j]=='#'){dame=true;}
		zyun[mae[j]-'a'][now[j]-'a']=1;
		
		
		mae=now;
		
	}
	for(i=0;i<26;i++){
		for(j=0;j<26;j++){
			if(zyun[i][j]){zi[j]++;}
		}
	}
	if(dame){cout<<"no"<<endl;return true;}
	queue<int>que;
	for(i=0;i<26;i++){
		if(zi[i]==0){que.push(i);}
	}
	while(que.size()){
		int t=que.front();que.pop();
		for(j=0;j<26;j++){
			if(zyun[t][j]){
				
				zi[j]--;
				if(zi[j]==0){que.push(j);}
			}
		}
	}
	for(i=0;i<26;i++){
		if(zi[i]>0){cout<<"no"<<endl;return true;}
	}
	cout<<"yes"<<endl;
	return true;
}

int main(void){
	while(solve()){}
	return 0;
}
