#include <bits/stdc++.h>
using namespace std;
long long sum(vector<long long> v) {
  long long init = 0;
  return accumulate(v.begin(), v.end(), init);
}
long long sum(vector<pair<long long, long long> > v, int check) {
  long long init = 0;
  if (check == 1)
    return accumulate(
        v.begin(), v.end(), init,
        [](long long a, pair<long long, long long> b) { return a + b.first; });
  else
    return accumulate(
        v.begin(), v.end(), init,
        [](long long a, pair<long long, long long> b) { return a + b.second; });
}
vector<long long> prsum(vector<long long> v) {
  vector<long long> temp(v.size());
  for (int i = 0; i < v.size(); i++) {
    if (i > 0)
      temp[i] = temp[i - 1] + v[i];
    else
      temp[i] = v[i];
  }
  return temp;
}
const long long mod = 1000000007;
int main(int argc, char const *argv[]) {
  int n;
  cin >> n;
  list<string> myList;
  vector<string> v(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
  }
  sort(v.begin(), v.end(), [](string &s1, string &s2) {
    if (s1.size() < s2.size())
      return true;
    else if (s1 == s2)
      return s1 < s2;
    else
      return false;
  });
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < i; j++) {
      if (v[i].find(v[j]) == string::npos) {
        cout << "NO\n";
        return 0;
      }
    }
  }
  cout << "YES\n";
  for (auto it : v) {
    cout << it << endl;
  }
}

### Prompt

Your task is to create 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;
long long sum(vector<long long> v) {
  long long init = 0;
  return accumulate(v.begin(), v.end(), init);
}
long long sum(vector<pair<long long, long long> > v, int check) {
  long long init = 0;
  if (check == 1)
    return accumulate(
        v.begin(), v.end(), init,
        [](long long a, pair<long long, long long> b) { return a + b.first; });
  else
    return accumulate(
        v.begin(), v.end(), init,
        [](long long a, pair<long long, long long> b) { return a + b.second; });
}
vector<long long> prsum(vector<long long> v) {
  vector<long long> temp(v.size());
  for (int i = 0; i < v.size(); i++) {
    if (i > 0)
      temp[i] = temp[i - 1] + v[i];
    else
      temp[i] = v[i];
  }
  return temp;
}
const long long mod = 1000000007;
int main(int argc, char const *argv[]) {
  int n;
  cin >> n;
  list<string> myList;
  vector<string> v(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
  }
  sort(v.begin(), v.end(), [](string &s1, string &s2) {
    if (s1.size() < s2.size())
      return true;
    else if (s1 == s2)
      return s1 < s2;
    else
      return false;
  });
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < i; j++) {
      if (v[i].find(v[j]) == string::npos) {
        cout << "NO\n";
        return 0;
      }
    }
  }
  cout << "YES\n";
  for (auto it : v) {
    cout << it << endl;
  }
}

```