#include <bits/stdc++.h>
using namespace std;
int dirx[] = {1, 0, -1, 0};
int diry[] = {0, 1, 0, -1};
vector<string> v;
map<int, int> mp;
map<string, int> chk;
map<string, int> ase;
vector<pair<int, int> > vv;
map<string, int> global;
vector<string> q;
void fnc(string ans, int idx) {
  string p;
  chk.clear();
  for (int i = 1; i <= ans.size(); i++) {
    for (int j = 0; j <= ans.size() - i; j++) {
      p.clear();
      p = ans.substr(j, i);
      if (!chk[p] && ase[p]) {
        chk[p] = 1;
        mp[idx]++;
      }
    }
  }
  vv.push_back(make_pair(mp[idx], idx));
}
bool check(string ans, int idx) {
  string p;
  global.clear();
  for (int i = 1; i <= ans.size(); i++) {
    for (int j = 0; j <= ans.size() - i; j++) {
      p.clear();
      p = ans.substr(j, i);
      global[p] = 1;
    }
  }
  for (int i = 0; i < idx; i++) {
    if (!global[q[i]]) return false;
  }
  return true;
}
int main() {
  string str;
  string ans;
  int a, ln = 0;
  cin >> a;
  for (int i = 0; i < a; i++) {
    cin >> str;
    v.push_back(str);
    ase[str] = 1;
  }
  for (int i = 0; i < a; i++) {
    fnc(v[i], i);
  }
  sort(vv.begin(), vv.end());
  for (int i = 0; i < vv.size(); i++) {
    if (i == 0) {
      q.push_back(v[vv[i].second]);
      continue;
    } else {
      if (!check(v[vv[i].second], i)) {
        cout << "NO" << endl;
        return 0;
      } else
        q.push_back(v[vv[i].second]);
    }
  }
  cout << "YES" << endl;
  for (int i = 0; i < q.size(); i++) {
    cout << q[i] << endl;
  }
}

### Prompt

Please provide a Cpp coded solution 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;
int dirx[] = {1, 0, -1, 0};
int diry[] = {0, 1, 0, -1};
vector<string> v;
map<int, int> mp;
map<string, int> chk;
map<string, int> ase;
vector<pair<int, int> > vv;
map<string, int> global;
vector<string> q;
void fnc(string ans, int idx) {
  string p;
  chk.clear();
  for (int i = 1; i <= ans.size(); i++) {
    for (int j = 0; j <= ans.size() - i; j++) {
      p.clear();
      p = ans.substr(j, i);
      if (!chk[p] && ase[p]) {
        chk[p] = 1;
        mp[idx]++;
      }
    }
  }
  vv.push_back(make_pair(mp[idx], idx));
}
bool check(string ans, int idx) {
  string p;
  global.clear();
  for (int i = 1; i <= ans.size(); i++) {
    for (int j = 0; j <= ans.size() - i; j++) {
      p.clear();
      p = ans.substr(j, i);
      global[p] = 1;
    }
  }
  for (int i = 0; i < idx; i++) {
    if (!global[q[i]]) return false;
  }
  return true;
}
int main() {
  string str;
  string ans;
  int a, ln = 0;
  cin >> a;
  for (int i = 0; i < a; i++) {
    cin >> str;
    v.push_back(str);
    ase[str] = 1;
  }
  for (int i = 0; i < a; i++) {
    fnc(v[i], i);
  }
  sort(vv.begin(), vv.end());
  for (int i = 0; i < vv.size(); i++) {
    if (i == 0) {
      q.push_back(v[vv[i].second]);
      continue;
    } else {
      if (!check(v[vv[i].second], i)) {
        cout << "NO" << endl;
        return 0;
      } else
        q.push_back(v[vv[i].second]);
    }
  }
  cout << "YES" << endl;
  for (int i = 0; i < q.size(); i++) {
    cout << q[i] << endl;
  }
}

```