#include <bits/stdc++.h>
using namespace std;
int fx[] = {1, -1, 0, 0};
int fy[] = {0, 0, 1, -1};
vector<pair<int, string> > pq;
int main() {
  int N;
  cin >> N;
  for (int i = 1; i <= N; i++) {
    string str;
    cin >> str;
    pq.push_back(pair<int, string>(str.size(), str));
  }
  vector<string> v;
  vector<string> v2;
  sort(pq.begin(), pq.end());
  for (int i = 0; i < pq.size(); i++) {
    string str = pq[i].second;
    int f1 = 0, f2 = 0, f3 = 0, f4 = 0;
    for (int j = i + 1; j <= pq.size() - 1; j++) {
      string str2;
      str2 = pq[j].second;
      f4 = 0;
      for (int k = 0; k < str2.size(); k++) {
        if (f4) break;
        if (str2[k] == str[0]) {
          for (int x = 1; x <= str.size() - 1; x++) {
            if (str2[k + x] != str[x]) break;
            if (x == str.size() - 1) f4 = 1;
          }
          if (str.size() == 1) f4 = 1;
        }
      }
      if (!f4) break;
      if (j == pq.size() - 1) {
        v2.push_back(str);
      }
    }
  }
  if (v2.size() < N - 1) {
    cout << "NO\n";
    return 0;
  }
  cout << "YES\n";
  for (int i = 0; i < v2.size(); i++) cout << v2[i] << "\n";
  cout << pq[pq.size() - 1].second << "\n";
}

### Prompt

Create a solution in CPP for 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;
int fx[] = {1, -1, 0, 0};
int fy[] = {0, 0, 1, -1};
vector<pair<int, string> > pq;
int main() {
  int N;
  cin >> N;
  for (int i = 1; i <= N; i++) {
    string str;
    cin >> str;
    pq.push_back(pair<int, string>(str.size(), str));
  }
  vector<string> v;
  vector<string> v2;
  sort(pq.begin(), pq.end());
  for (int i = 0; i < pq.size(); i++) {
    string str = pq[i].second;
    int f1 = 0, f2 = 0, f3 = 0, f4 = 0;
    for (int j = i + 1; j <= pq.size() - 1; j++) {
      string str2;
      str2 = pq[j].second;
      f4 = 0;
      for (int k = 0; k < str2.size(); k++) {
        if (f4) break;
        if (str2[k] == str[0]) {
          for (int x = 1; x <= str.size() - 1; x++) {
            if (str2[k + x] != str[x]) break;
            if (x == str.size() - 1) f4 = 1;
          }
          if (str.size() == 1) f4 = 1;
        }
      }
      if (!f4) break;
      if (j == pq.size() - 1) {
        v2.push_back(str);
      }
    }
  }
  if (v2.size() < N - 1) {
    cout << "NO\n";
    return 0;
  }
  cout << "YES\n";
  for (int i = 0; i < v2.size(); i++) cout << v2[i] << "\n";
  cout << pq[pq.size() - 1].second << "\n";
}

```