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;
map<pair<pair<long long int, long long int>,
         pair<long long int, long long int> >,
    long long int>
    pqr;
map<pair<long long int, long long int>, long long int> xyz;
vector<pair<int, pair<int, int> > > a;
vector<pair<pair<int, int>, pair<int, int> > > b;
vector<pair<int, int> > c;
bool compare(string &x, string &y) { return x.size() < y.size(); }
int main() {
  int n, i;
  cin >> n;
  vector<string> arr(n);
  for (i = 0; i < n; i++) {
    cin >> arr[i];
  }
  sort(arr.begin(), arr.end(), compare);
  for (i = 0; i < n - 1; i++) {
    if (arr[i + 1].find(arr[i]) == string::npos) {
      cout << "NO";
      return 0;
    }
  }
  cout << "YES" << endl;
  for (auto it : arr) {
    cout << it << endl;
  }
  return 0;
}
