#include <bits/stdc++.h>
using namespace std;
int main() {
  vector<pair<int, string> > data;
  int n;
  string kata;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> kata;
    data.push_back(make_pair(kata.length(), kata));
  }
  sort(data.begin(), data.end());
  for (int i = 0; i < n - 1; i++) {
    for (int j = i + 1; j < n; j++) {
      if (data[j].second.find(data[i].second) == string::npos)
        return cout << "NO" << endl, 0;
    }
  }
  cout << "YES" << endl;
  for (int i = 0; i < data.size(); i++) cout << data[i].second << endl;
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

You are given n strings. Each string consists of lowercase English letters. Rearrange (reorder) the given strings in such a way that for every string, all strings that are placed before it are its substrings.

String a is a substring of string b if it is possible to choose several consecutive letters in b in such a way that they form a. For example, string "for" is contained as a substring in strings "codeforces", "for" and "therefore", but is not contained as a substring in strings "four", "fofo" and "rof".

Input

The first line contains an integer n (1 ≤ n ≤ 100) — the number of strings.

The next n lines contain the given strings. The number of letters in each string is from 1 to 100, inclusive. Each string consists of lowercase English letters.

Some strings might be equal.

Output

If it is impossible to reorder n given strings in required order, print "NO" (without quotes).

Otherwise print "YES" (without quotes) and n given strings in required order.

Examples

Input

5
a
aba
abacaba
ba
aba


Output

YES
a
ba
aba
aba
abacaba


Input

5
a
abacaba
ba
aba
abab


Output

NO


Input

3
qwerty
qwerty
qwerty


Output

YES
qwerty
qwerty
qwerty

Note

In the second example you cannot reorder the strings because the string "abab" is not a substring of the string "abacaba".

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  vector<pair<int, string> > data;
  int n;
  string kata;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> kata;
    data.push_back(make_pair(kata.length(), kata));
  }
  sort(data.begin(), data.end());
  for (int i = 0; i < n - 1; i++) {
    for (int j = i + 1; j < n; j++) {
      if (data[j].second.find(data[i].second) == string::npos)
        return cout << "NO" << endl, 0;
    }
  }
  cout << "YES" << endl;
  for (int i = 0; i < data.size(); i++) cout << data[i].second << endl;
  return 0;
}

```