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;
const long long mod = 1000000007;
const long long inf = 1000000000009;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int n{};
  vector<pair<int, string>> v;
  cin >> n;
  int mn = 1e+5;
  int pos{-1};
  for (int i = 0, sz = (int)n; i < sz; i++) {
    string s;
    cin >> s;
    v.push_back({s.size(), s});
  }
  sort((v).begin(), (v).end());
  int f{};
  for (int i = 0, sz = (int)n - 1; i < sz; i++) {
    if (v[i + 1].second.find(v[i].second) == -1) {
      f = 1;
      break;
    }
  }
  if (f)
    cout << "NO"
         << "\n";
  else {
    cout << "YES"
         << "\n";
    for (int i = 0, sz = (int)n; i < sz; i++) cout << v[i].second << "\n";
  }
  return 0;
}
