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;
bool vowel(char a) {
  a = toupper(a);
  if ((a == 'A') || (a == 'E') || (a == 'I') || (a == 'O') || (a == 'U'))
    return 1;
  return 0;
}
long long int n, m, q, a, b, c, sum = 0, cnt = 0;
long long int arr[1000055];
map<long long int, bool> mp;
string s;
vector<string> v;
bool cmp(string a, string b) { return a.size() < b.size(); }
bool isSub(string a, string b) {
  int cnt;
  for (int i = 0; i < b.size(); i++) {
    int tp = i;
    for (int j = 0; j < a.size(); j++) {
      if (a[j] != b[tp]) break;
      if (j == a.size() - 1) return true;
      if (tp == b.size() - 1) return false;
      tp++;
    }
  }
  return false;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> s;
    v.push_back(s);
  }
  sort(v.begin(), v.end(), cmp);
  for (int i = 1; i <= n - 1; i++) {
    if (!isSub(v[i - 1], v[i])) {
      cout << "NO" << endl;
      return 0;
    }
  }
  cout << "YES" << endl;
  for (int i = 0; i < n; i++) cout << v[i] << endl;
}
