#include <bits/stdc++.h>
using namespace std;
long long power(long long b, long long e, long long m) {
  if (e == 0) return 1;
  if (e & 1) return b * power(b * b % m, e / 2, m) % m;
  return power(b * b % m, e / 2, m);
}
long long power(long long b, long long e) {
  if (e == 0) return 1;
  if (e & 1) return b * power(b * b, e / 2);
  return power(b * b, e / 2);
}
int isSubstring(string s1, string s2) {
  int M = s1.length();
  int N = s2.length();
  for (int i = 0; i <= N - M; i++) {
    int j;
    for (j = 0; j < M; j++)
      if (s2[i + j] != s1[j]) break;
    if (j == M) return 1;
  }
  return 0;
}
void solve() {
  int n;
  cin >> n;
  vector<pair<int, string> > v;
  for (int i = 0; i < n; i++) {
    string a;
    cin >> a;
    int x = a.size();
    v.push_back({x, a});
  }
  sort((v).begin(), (v).end());
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) {
      if (!isSubstring(v[i].second, v[j].second)) {
        cout << "NO"
             << "\n";
        return;
      }
    }
  }
  cout << "YES"
       << "\n";
  for (int i = 0; i < n; i++) cout << v[i].second << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) solve();
  return 0;
}

### Prompt

Please create a solution in CPP 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 power(long long b, long long e, long long m) {
  if (e == 0) return 1;
  if (e & 1) return b * power(b * b % m, e / 2, m) % m;
  return power(b * b % m, e / 2, m);
}
long long power(long long b, long long e) {
  if (e == 0) return 1;
  if (e & 1) return b * power(b * b, e / 2);
  return power(b * b, e / 2);
}
int isSubstring(string s1, string s2) {
  int M = s1.length();
  int N = s2.length();
  for (int i = 0; i <= N - M; i++) {
    int j;
    for (j = 0; j < M; j++)
      if (s2[i + j] != s1[j]) break;
    if (j == M) return 1;
  }
  return 0;
}
void solve() {
  int n;
  cin >> n;
  vector<pair<int, string> > v;
  for (int i = 0; i < n; i++) {
    string a;
    cin >> a;
    int x = a.size();
    v.push_back({x, a});
  }
  sort((v).begin(), (v).end());
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) {
      if (!isSubstring(v[i].second, v[j].second)) {
        cout << "NO"
             << "\n";
        return;
      }
    }
  }
  cout << "YES"
       << "\n";
  for (int i = 0; i < n; i++) cout << v[i].second << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) solve();
  return 0;
}

```