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<bits/stdc++.h>
using namespace std;
using ll=long long;
using ld=long double;
using P=pair<ll,ll>;
#define MOD 1000000007ll
#define INF 1000000000ll
#define EPS 1e-10
#define FOR(i,n,m) for(ll i=n;i<(ll)m;i++)
#define REP(i,n) FOR(i,0,n)
#define DUMP(a) REP(d,a.size()){cout<<a[d];if(d!=a.size()-1)cout<<" ";else cout<<endl;}
#define ALL(v) v.begin(),v.end()
#define UNIQUE(v) sort(v.begin(),v.end());v.erase(unique(v.begin(),v.end()),v.end());
#define pb push_back

ll n;
vector<string> s;
vector<vector<ll>> g(26);
vector<bool> sorted;

bool dfs(ll k) {
	bool upd=false;
	bool ret=true;
	REP(i,n-1) {
		if(!sorted[i]&&s[i].size()>k&&s[i+1].size()==k) ret=false;
		if(s[i].size()<=k||s[i+1].size()<=k) continue;
		upd=true;
		if(!sorted[i]&&s[i][k]!=s[i+1][k]) {
			sorted[i]=true;
			g[s[i][k]-'a'].pb(s[i+1][k]-'a');
		}
	}
	if(upd) ret=ret&dfs(k+1);
	return ret;
}

bool dfs2(ll p, ll par, ll tgt, vector<bool>& used) {
	used[p]=true;
	bool ret=true;
	REP(i,g[p].size()) {
		if(g[p][i]==tgt) return false;
		if(g[p][i]==par) continue;
		if(used[g[p][i]]) continue;
		else ret=ret&dfs2(g[p][i],p,tgt,used);
	}
	return ret;
}

bool is_dag() {
	REP(i,26) {
		if(g[i].empty()) continue;
		sort(ALL(g[i]));
		g[i].erase(unique(ALL(g[i])),g[i].end());
	}
	REP(i,26) {
		vector<bool> used(26,false);
		if(!dfs2(i,-1,i,used)) return false;
	}
	return true;
}

string solve() {
	s.assign(n,"");
	g.assign(26,vector<ll>(0));
	sorted.assign(n-1,false);
	REP(i,n) cin>>s[i];
	if(!dfs(0)) return "no";
	if(is_dag()) return "yes";
	else return "no";
}

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	while(1) {
		cin>>n;
		if(n==0) break;
		cout<<solve()<<endl;
	}
}

