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;
int main() {
  int n, t = 0, t1 = 0, lmax = 0;
  cin >> n;
  string s[n], s1[n], smax;
  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].length() > s[j].length()) {
        string temp = s[i];
        s[i] = s[j];
        s[j] = temp;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    t = 0;
    for (int j = i + 1; j < n; j++) {
      if (s[j].find(s[i]) == string ::npos) {
        t++;
      }
    }
    if (t == 0) {
      t1++;
    } else
      break;
  }
  if (t1 == n) {
    cout << "YES\n";
    for (int i = 0; i < n; i++) cout << s[i] << endl;
  } else
    cout << "NO\n";
}
