#include <bits/stdc++.h>
using namespace std;
vector<pair<int, string>> v;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    int n = s.size();
    v.push_back({n, s});
  }
  sort(v.begin(), v.end());
  for (int i = 1; i < v.size(); ++i) {
    if (v[i].second.find(v[i - 1].second) == ::string::npos) {
      cout << "NO\n";
      return 0;
    }
  }
  cout << "YES\n";
  for (int i = 0; i < v.size(); ++i) cout << v[i].second << '\n';
  return 0;
}

### Prompt

Generate a CPP solution to the following problem:

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;
vector<pair<int, string>> v;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    int n = s.size();
    v.push_back({n, s});
  }
  sort(v.begin(), v.end());
  for (int i = 1; i < v.size(); ++i) {
    if (v[i].second.find(v[i - 1].second) == ::string::npos) {
      cout << "NO\n";
      return 0;
    }
  }
  cout << "YES\n";
  for (int i = 0; i < v.size(); ++i) cout << v[i].second << '\n';
  return 0;
}

```