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 <algorithm>
#include <vector>
#include <string>

using namespace std;

using lint = long long;
using ldouble = long double;

bool solve() {
    int N;
    cin >> N;
    if (N == 0) return false;

    vector<string> S(N);
    for (auto& s : S) cin >> s;

    vector<vector<bool>> order(26, vector<bool>(26, false));
    // order[a][b] => a < b

    for (int i = 0; i < N - 1; ++i) {
        bool diff = false;
        for (int j = 0; j < min(S[i].length(), S[i + 1].length()); ++j) {
            if (S[i][j] != S[i + 1][j]) {
                order[S[i][j] - 'a'][S[i + 1][j] - 'a'] = true;
                diff = true;
                break;
            }
        }

        if (!diff && S[i].length() > S[i + 1].length()) {
            cout << "no" << endl;
            return true;
        }
    }

    for (int j = 0; j < 26; ++j) {
        for (int i = 0; i < 26; ++i) {
            for (int k = 0; k < 26; ++k) {
                if (order[i][j] && order[j][k]) order[i][k] = true;
            }
        }
    }

    for (int i = 0; i < 26; ++i) {
        if (order[i][i]) {
            cout << "no" << endl;
            return true;
        }
    }
    cout << "yes" << endl;
    return true;
}

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

