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;

//#define int long long
#define reps(i,s,n) for(int (i)=(s);(i)<(n);++(i))
#define rep(i,n) reps(i,0,n)
#define rept(i,n) rep(i,(n)+1)
#define repst(i,s,n) reps(i,s,(n)+1)
#define reprt(i,n,t) for(int (i)=(n);(i)>=(t);--(i))
#define repr(i,n) reprt(i,n,0)
#define each(itr,v) for(auto &(itr):(v))
#define all(c) (c).begin(),(c).end()
#define rall(c) (c).rbegin(),(c).rend()
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define tmax(x,y,z) max(x,max(y,z))
#define tmin(x,y,z) min(x,min(y,z))
#define chmin(x,y) x=min(x,y)
#define chmax(x,y) x=max(x,y)
#define ln '\n'
#define bln(i,n) (((i)==(n)-1)?'\n':' ')
#define dbg(x) cout<<#x" = "<<(x)<<ln<<flush
#define dbga(x,n) {cout<<#x" : ";for(int (i)=0;i<(n);++i){cout<<((x)[i])<<(i==((n)-1)?'\n':' ')<<flush;}}
#define zero(a) memset(a,0,sizeof(a))
#define unq(a) sort(all(a)),a.erase(unique(all(a)),a.end())

typedef complex<double> P;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<string> vst;
typedef vector<pii> vpii;
typedef vector<pll> vpll;
typedef vector<vector<int> > mat;

const ll inf = (ll)1e9+10;
const ll linf = (ll)1e18+10;
const ll mod = (ll)(1e9+7);
const int dx[] = {0, 1, 0, -1};
const int dy[] = {1, 0, -1, 0};
const int ddx[] = {0, 1, 1, 1, 0, -1, -1, -1};
const int ddy[] = {1, 1, 0, -1, -1, -1, 0, 1};
const double eps = 1e-10;

ll mop(ll a,ll b,ll m=mod) {ll r=1;a%=m;for(;b;b>>=1){if(b&1)r=r*a%m;a=a*a%m;}return r;}
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b) {return a*b/gcd(a,b);}
bool ool(int x,int y,int h,int w) {return((x<0)||(h<=x)||(y<0)||(w<=y));}
bool deq(double a,double b) {return abs(a-b)<eps;}

struct oreno_initializer {
	oreno_initializer() {
		cin.tie(0);
		ios::sync_with_stdio(0);
	}
} oreno_initializer;

// ━━━━☆・‥…━━━☆・‥…━━━☆・‥…━━━☆・‥…━━━☆・‥…━━━☆・‥…━━━☆・‥…━━━☆・‥…━━━☆・‥…
// .｡.:( ^ω^)・ﾟ＋.｡.:( ^ω^)・ﾟ＋.｡.:( ^ω^)・ﾟ＋.｡.:( ^ω^)・ﾟ＋.｡.:( ^ω^)・ﾟ＋
// ・‥…━━━☆・‥…━━━☆・‥…━━━☆・‥…━━━☆・‥…━━━☆・‥…━━━☆・‥…━━━☆・‥…━━━☆・‥…━━━☆・

// ex2だとまず先頭1文字だけを見るとb<aが確定
// 上2つはokだけど下2つでprefixの条件を破ってるからダメ
// こんな感じで条件を全部列挙して矛盾があったらno
// ex4の条件は b<c(1個目と2個目より) と c<b(3個目と4個目) で矛盾してるからダメ
// a<bならa->bに辺をつなげるとかして出来たグラフに閉路があれば矛盾してるとわかる

int n;
string s[500];
vi e[26];
bool vis[26], hei;

void rec(int v, int p) {
	vis[v] = 1;
	rep(i,e[v].size()) {
		if (e[v][i]==p) hei = 1;
		else if (!vis[e[v][i]]) rec(e[v][i],p);
	}
}

signed main() {
	while (1) {
		cin >> n;
		if (n==0) break;
		rep(i,26) e[i].clear();
		bool dame = 0;
		rep(i,n) {
			cin >> s[i];
			rep(j,i) if (s[j].size()>s[i].size() && s[j].substr(0,s[i].size())==s[i]) dame = 1; // prefix
		}
		if (dame) {
			cout << "no" << ln;
			continue;
		}
		// 条件追加
		rep(i,n) reps(j,i+1,n) {
			if (s[j].size()>=s[i].size() && s[j].substr(0,s[i].size())==s[i]) continue; // prefix
			rep(k,min(s[i].size(),s[j].size())) if (s[i][k]!=s[j][k]) {
				e[s[i][k]-'a'].pb(s[j][k]-'a');
				break;
			}
		}
		// 閉路検出
		hei = 0;
		rep(i,26) zero(vis), rec(i,i);
		cout << (hei ? "no" : "yes") << ln;
	}
}
