#include <bits/stdc++.h>
using ll = long long;
ll OO = 2e18, mod = 1e9 + 7;
const int oo = 0x3f3f3f3f;
using namespace std;
ll gcd(ll a, ll b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
ll lcm(ll a, ll b) { return a * b / gcd(a, b); }
ll pw(ll base, ll os) {
  if (os == 0) return 1;
  ll z = pw(base, os / 2);
  z *= z;
  return (os % 2 ? z * base : z);
}
ll pwm(ll base, ll os) {
  if (os == 0) return 1;
  ll z = pwm(base, os / 2);
  z = ((z % mod) * (z % mod)) % mod;
  return (os % 2 ? ((z % mod) * (base % mod)) % mod : z % mod);
}
bool isprime(ll x) {
  for (ll i = 2; i * i <= x; i += 1 + (1 & i)) {
    if (x % i == 0) return false;
  }
  return x > 1;
}
int XX[] = {-1, 0, 1, 0, -1, 1, 1, -1};
int YY[] = {0, 1, 0, -1, 1, 1, -1, -1};
const int N = 1000;
bool found(string &src, string &dst) {
  int n = src.size(), m = dst.size();
  for (int j = 0; j < m; j++) {
    bool ok = 1;
    for (int i = 0; i < n; i++) {
      if (src[i] != dst[j + i]) {
        ok = 0;
        break;
      }
    }
    if (ok) return 1;
  }
  return false;
}
void run_case() {
  int n;
  cin >> n;
  string s;
  vector<pair<int, string> > a;
  while (n--) {
    cin >> s;
    a.push_back({s.size(), s});
  }
  sort(a.begin(), a.end());
  n = a.size();
  for (int i = 0; i < n; i++) {
    s = a[i].second;
    for (int j = i + 1; j < n; j++) {
      if (!found(s, a[j].second)) {
        cout << "NO\n";
        return;
      }
    }
  }
  cout << "YES\n";
  for (auto vis : a) cout << vis.second << "\n";
}
int main() {
  ios::sync_with_stdio(0), cout.tie(0), cin.tie(0);
  int tc = 1;
  while (tc--) run_case();
}

### Prompt

Generate 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 ll = long long;
ll OO = 2e18, mod = 1e9 + 7;
const int oo = 0x3f3f3f3f;
using namespace std;
ll gcd(ll a, ll b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
ll lcm(ll a, ll b) { return a * b / gcd(a, b); }
ll pw(ll base, ll os) {
  if (os == 0) return 1;
  ll z = pw(base, os / 2);
  z *= z;
  return (os % 2 ? z * base : z);
}
ll pwm(ll base, ll os) {
  if (os == 0) return 1;
  ll z = pwm(base, os / 2);
  z = ((z % mod) * (z % mod)) % mod;
  return (os % 2 ? ((z % mod) * (base % mod)) % mod : z % mod);
}
bool isprime(ll x) {
  for (ll i = 2; i * i <= x; i += 1 + (1 & i)) {
    if (x % i == 0) return false;
  }
  return x > 1;
}
int XX[] = {-1, 0, 1, 0, -1, 1, 1, -1};
int YY[] = {0, 1, 0, -1, 1, 1, -1, -1};
const int N = 1000;
bool found(string &src, string &dst) {
  int n = src.size(), m = dst.size();
  for (int j = 0; j < m; j++) {
    bool ok = 1;
    for (int i = 0; i < n; i++) {
      if (src[i] != dst[j + i]) {
        ok = 0;
        break;
      }
    }
    if (ok) return 1;
  }
  return false;
}
void run_case() {
  int n;
  cin >> n;
  string s;
  vector<pair<int, string> > a;
  while (n--) {
    cin >> s;
    a.push_back({s.size(), s});
  }
  sort(a.begin(), a.end());
  n = a.size();
  for (int i = 0; i < n; i++) {
    s = a[i].second;
    for (int j = i + 1; j < n; j++) {
      if (!found(s, a[j].second)) {
        cout << "NO\n";
        return;
      }
    }
  }
  cout << "YES\n";
  for (auto vis : a) cout << vis.second << "\n";
}
int main() {
  ios::sync_with_stdio(0), cout.tie(0), cin.tie(0);
  int tc = 1;
  while (tc--) run_case();
}

```