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".
#include <bits/stdc++.h>
using namespace std;
void solve() {
  string s[1000], tmp;
  int n, len, flag;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> s[i];
  for (int i = 0; i < n; i++)
    for (int j = i + 1; j < n; j++)
      if (s[i].size() > s[j].size()) swap(s[i], s[j]);
  tmp = s[0];
  len = s[0].size();
  for (int i = 0; i < n; i++) {
    flag = true;
    for (int j = 0; j < s[i].size() - len + 1; j++) {
      if (s[i].substr(j, len) == tmp) {
        flag = false;
      }
    }
    if (flag) {
      cout << "NO";
      return;
    }
    tmp = s[i];
    len = s[i].size();
  }
  cout << "YES" << endl;
  for (int i = 0; i < n; i++) cout << s[i] << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  solve();
}
