#include <bits/stdc++.h>
using namespace std;
const int INF = 0x7badface;
const long long int YNF = 0x7facedeafcafebed;
const string NL("\n");
const string SEP = string(30, '-') + NL;
template <class T>
istream& operator>>(istream& r, vector<T>& v) {
  for (auto& x : v) r >> x;
  return r;
}
template <class S, class T>
istream& operator>>(istream& r, pair<S, T>& p) {
  r >> p.first >> p.second;
  return r;
}
template <class T>
ostream& operator<<(ostream& r, const vector<T>& v) {
  for (const auto& x : v) r << x << NL;
  return r;
}
template <class T>
ostream& operator<<(ostream& r, const vector<vector<T>>& v) {
  r << NL << SEP;
  for (const auto& x : v) r << x << NL;
  r << SEP;
  return r;
}
template <class S, class T>
ostream& operator<<(ostream& r, const pair<S, T>& p) {
  r << p.first << NL << p.second;
  return r;
}
bool Cmp(const string& a, const string& b) { return a.size() < b.size(); }
pair<string, vector<string>> SexyLady() {
  int n;
  cin >> n;
  vector<string> v(n);
  cin >> v;
  sort(v.begin(), v.end(), Cmp);
  for (int i = 1; i < v.size(); ++i) {
    if (v[i].find(v[i - 1]) == string::npos) {
      return {"NO", {}};
    }
  }
  return {"YES", v};
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout << NL;
  int t = 1;
  while (t--) {
    cout << SexyLady() << NL;
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 INF = 0x7badface;
const long long int YNF = 0x7facedeafcafebed;
const string NL("\n");
const string SEP = string(30, '-') + NL;
template <class T>
istream& operator>>(istream& r, vector<T>& v) {
  for (auto& x : v) r >> x;
  return r;
}
template <class S, class T>
istream& operator>>(istream& r, pair<S, T>& p) {
  r >> p.first >> p.second;
  return r;
}
template <class T>
ostream& operator<<(ostream& r, const vector<T>& v) {
  for (const auto& x : v) r << x << NL;
  return r;
}
template <class T>
ostream& operator<<(ostream& r, const vector<vector<T>>& v) {
  r << NL << SEP;
  for (const auto& x : v) r << x << NL;
  r << SEP;
  return r;
}
template <class S, class T>
ostream& operator<<(ostream& r, const pair<S, T>& p) {
  r << p.first << NL << p.second;
  return r;
}
bool Cmp(const string& a, const string& b) { return a.size() < b.size(); }
pair<string, vector<string>> SexyLady() {
  int n;
  cin >> n;
  vector<string> v(n);
  cin >> v;
  sort(v.begin(), v.end(), Cmp);
  for (int i = 1; i < v.size(); ++i) {
    if (v[i].find(v[i - 1]) == string::npos) {
      return {"NO", {}};
    }
  }
  return {"YES", v};
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout << NL;
  int t = 1;
  while (t--) {
    cout << SexyLady() << NL;
  }
  return 0;
}

```