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 n;
bool check = 1;
vector<string> vec;
vector<string> tes[101];
int main() {
  cin >> n;
  for (int i = 0; i < n; ++i) {
    string tmp;
    cin >> tmp;
    tes[tmp.size()].push_back(tmp);
  }
  for (int i = 0; i < 101; i++) {
    if (tes[i].size() != 0) {
      for (int j = 0; j < tes[i].size(); j++) vec.push_back(tes[i][j]);
    }
  }
  for (int i = 0; i < n - 1; ++i) {
    if (vec[i + 1].find(vec[i]) == string::npos) {
      check = 0;
      break;
    }
  }
  if (check) {
    cout << "YES\n";
    for (int i = 0; i < n; ++i) cout << vec[i] << "\n";
  } else {
    cout << "NO\n";
  }
}
