#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const long long LL_INF = 0x3f3f3f3f3f3f3f3f;
const long double PI = acos(-1);
int subS(string a, string b) {
  for (int i = 0; i < int((b).size()) - int((a).size()) + 1; i++) {
    string s = b.substr(i, int((a).size()));
    if (a == s) return 1;
  }
  return 0;
}
bitset<101> sub[101];
bitset<101> vis;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  cin >> n;
  vector<string> tab(n);
  for (auto& i : tab) cin >> i;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i != j) {
        sub[i][j] = subS(tab[i], tab[j]);
      }
    }
  }
  vector<int> odp;
  for (int T = 0; T < n; T++) {
    bool dodal = false;
    for (int i = 0; i < n; i++) {
      if (vis[i]) continue;
      int ile = 0;
      for (int j = 0; j < n; j++) {
        if (!vis[j])
          if (sub[i][j]) ile++;
      }
      if (ile == n - T - 1) {
        odp.push_back(i);
        vis[i] = true;
        dodal = true;
        break;
      }
    }
    if (!dodal) break;
  }
  if (int((odp).size()) != n) {
    cout << "NO";
  } else {
    cout << "YES\n";
    for (auto& i : odp) {
      cout << tab[i] << "\n";
    }
  }
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

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".

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const long long LL_INF = 0x3f3f3f3f3f3f3f3f;
const long double PI = acos(-1);
int subS(string a, string b) {
  for (int i = 0; i < int((b).size()) - int((a).size()) + 1; i++) {
    string s = b.substr(i, int((a).size()));
    if (a == s) return 1;
  }
  return 0;
}
bitset<101> sub[101];
bitset<101> vis;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  cin >> n;
  vector<string> tab(n);
  for (auto& i : tab) cin >> i;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i != j) {
        sub[i][j] = subS(tab[i], tab[j]);
      }
    }
  }
  vector<int> odp;
  for (int T = 0; T < n; T++) {
    bool dodal = false;
    for (int i = 0; i < n; i++) {
      if (vis[i]) continue;
      int ile = 0;
      for (int j = 0; j < n; j++) {
        if (!vis[j])
          if (sub[i][j]) ile++;
      }
      if (ile == n - T - 1) {
        odp.push_back(i);
        vis[i] = true;
        dodal = true;
        break;
      }
    }
    if (!dodal) break;
  }
  if (int((odp).size()) != n) {
    cout << "NO";
  } else {
    cout << "YES\n";
    for (auto& i : odp) {
      cout << tab[i] << "\n";
    }
  }
  return 0;
}

```