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;
long long i, j, k;
int main() {
  int n;
  cin >> n;
  map<pair<int, string>, int> mp;
  for (i = 0; i < n; i++) {
    string s;
    cin >> s;
    int l = s.size();
    mp[make_pair(l, s)]++;
  }
  map<int, int> m2p;
  for (auto x : mp) {
    m2p[x.first.first]++;
    if (m2p[x.first.first] == 2) {
      cout << "NO";
      return 0;
    }
  }
  vector<string> v;
  for (auto x : mp) {
    while (x.second--) v.push_back(x.first.second);
  }
  for (i = 0; i < v.size() - 1; i++) {
    string s, su;
    s = v[i + 1];
    su = v[i];
    if (s.find(su) == -1) {
      cout << "NO";
      return 0;
    }
  }
  cout << "YES" << endl;
  for (i = 0; i < v.size(); i++) cout << v[i] << endl;
}
