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 <cstdio>
#include <string>
#include <cstring>
#include <deque>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <algorithm>
#include <map>
#include <set>
#include <complex>
#include <cmath>
#include <limits>
#include <cfloat>
#include <climits>
#include <ctime>
#include <cassert>
#include <numeric>
#include <functional>
#include <bitset>

using namespace std;
#define int long long int
const int INF = 1001001001001001LL;
const int MOD = 1000000007;

int di[] = {-1, 0, 1, 0, -1, 1, 1, -1};
int dj[] = {0, 1, 0, -1, 1, 1, -1, -1};

// トポソ
// O(V + E)
// [入力] 重みなし有効グラフ
// [出力] ソートされた頂点列(0-indexed) DAG じゃ無い場合は空列を返す
vector<int> toposort(vector<vector<int>> &g){
    int n = (int)g.size();
    vector<int> d(n, 0);     // 入次数
    for(int i = 0; i < n; i++){
        for(int j = 0; j < (int)g[i].size(); j++){
            d[g[i][j]]++;
        }
    }
    
    // 入次数が 0 の頂点を queue に入れる
    queue<int> que;
    for(int i = 0; i < n; i++){
        if(d[i] == 0) que.push(i);
    }

    vector<int> ret;
    while(!que.empty()){
        int cur = que.front();
        que.pop();
        ret.push_back(cur);

        for(int i = 0; i < (int)g[cur].size(); i++){
            int nxt = g[cur][i];
            d[nxt]--;
            if(d[nxt] == 0) que.push(nxt);
        }
    }

    // DAG じゃなかったら、頂点列が不足しているはず
    if(ret.size() == n) return ret;
    else return vector<int>();

}

signed main(){
    
    while(1){
        int n; cin >> n;
        if(!n) break;
        vector<string> vs(n);
        for(int i = 0; i < n; i++) cin >> vs[i];
        
        vector<vector<int>> g(26);
    
        // a -> b
        bool end = false;
        for(int i = 0; i < n - 1; i++){
            //cout << i << endl;
            int x = (int)vs[i].size();
            int y = (int)vs[i + 1].size();

            bool change = false;
            for(int si = 0; si < min(x, y); si++){
                if(vs[i][si] == vs[i + 1][si]) continue;
                
                change = true;
                int from = vs[i + 1][si] - 'a';
                int to = vs[i][si] - 'a';
                g[from].push_back(to);
                break;
            }

            if(!change){
                if(x > y){
                    cout << "no" << endl;
                    end = true;
                    break;
                }
            }
        }
        if(end) continue;
    
        
        // 閉路判定
        
        if(toposort(g).size() != 0) cout << "yes" << endl;
        else cout << "no" << endl;
    }
    return 0;
}

