#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
const int N = 200001;
set<char> vowel = {'A', 'O', 'Y', 'E', 'U', 'I', 'a', 'o', 'y', 'e', 'u', 'i'};
long long gcd(long long a, long long b) { return (b == 0 ? a : gcd(b, a % b)); }
bool issquare(long long w) { return trunc(sqrt(w)) * trunc(sqrt(w)) == w; }
int dx[] = {1, -1, 0, 0, 1, -1, 1, -1};
int dy[] = {0, 0, 1, -1, 1, 1, -1, -1};
int n, m, k, a, l, ans = 0, res = 0, posmin, posmax, x, y = 0, z = 0, xmax = 0,
                   ymax = 0;
bool vis[N], check = true;
char c, c1;
string ch, ch1, ch2;
queue<int> q;
priority_queue<int> pq;
map<int, set<int> > adj;
map<pair<int, int>, vector<pair<int, int> > > mapai;
map<int, set<int> > edje;
map<string, int> mastring;
map<char, int> machar;
unordered_map<int, int> mai;
set<char> setchar;
set<int> sx, sy, si;
vector<int> v, vv;
vector<pair<int, int> > vvv;
vector<string> vs;
bool comp(string sh, string sh1) { return sh.size() < sh1.size(); }
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> ch;
    vs.push_back(ch);
  }
  sort(vs.begin(), vs.end(), comp);
  for (int i = 1; i < n; i++) {
    if (vs[i].find(vs[i - 1]) >= vs[i].size()) {
      cout << "NO";
      return 0;
    }
  }
  cout << "YES" << endl;
  for (int i = 0; i < n; i++) {
    cout << vs[i] << endl;
  }
  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 MOD = 1000000007;
const int N = 200001;
set<char> vowel = {'A', 'O', 'Y', 'E', 'U', 'I', 'a', 'o', 'y', 'e', 'u', 'i'};
long long gcd(long long a, long long b) { return (b == 0 ? a : gcd(b, a % b)); }
bool issquare(long long w) { return trunc(sqrt(w)) * trunc(sqrt(w)) == w; }
int dx[] = {1, -1, 0, 0, 1, -1, 1, -1};
int dy[] = {0, 0, 1, -1, 1, 1, -1, -1};
int n, m, k, a, l, ans = 0, res = 0, posmin, posmax, x, y = 0, z = 0, xmax = 0,
                   ymax = 0;
bool vis[N], check = true;
char c, c1;
string ch, ch1, ch2;
queue<int> q;
priority_queue<int> pq;
map<int, set<int> > adj;
map<pair<int, int>, vector<pair<int, int> > > mapai;
map<int, set<int> > edje;
map<string, int> mastring;
map<char, int> machar;
unordered_map<int, int> mai;
set<char> setchar;
set<int> sx, sy, si;
vector<int> v, vv;
vector<pair<int, int> > vvv;
vector<string> vs;
bool comp(string sh, string sh1) { return sh.size() < sh1.size(); }
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> ch;
    vs.push_back(ch);
  }
  sort(vs.begin(), vs.end(), comp);
  for (int i = 1; i < n; i++) {
    if (vs[i].find(vs[i - 1]) >= vs[i].size()) {
      cout << "NO";
      return 0;
    }
  }
  cout << "YES" << endl;
  for (int i = 0; i < n; i++) {
    cout << vs[i] << endl;
  }
  return 0;
}

```