#include <bits/stdc++.h>
using namespace std;
const int MAXN = 100 + 10;
const int MAXL = 100 + 10;
char str[MAXN][MAXL];
vector<int> v[MAXN], results;
int n, times[MAXN], deg[MAXN];
bool check(int x, int y) {
  int n = strlen(str[x]), m = strlen(str[y]);
  for (int i = 0; i + n <= m; i++) {
    bool ok = true;
    for (int j = 0; j < n && ok; j++)
      if (str[x][j] != str[y][i + j]) ok = false;
    if (ok) return true;
  }
  return false;
}
void out() {
  results.clear();
  queue<int> q;
  for (; !q.empty(); q.pop())
    ;
  for (int i = 0; i < n; i++)
    if (!deg[i]) {
      if (!q.empty()) {
        puts("NO");
        return;
      }
      q.push(i);
    }
  for (; !q.empty();) {
    int u = q.front();
    q.pop();
    results.push_back(u);
    for (int i = 0; i < (signed)v[u].size(); i++) {
      if (!(--deg[v[u].at(i)])) {
        if (!q.empty()) {
          puts("NO");
          return;
        }
        q.push(v[u].at(i));
      }
    }
  }
  puts("YES");
  for (int i = 0; i < (signed)results.size(); i++)
    for (int j = 0; j < times[results.at(i)]; j++) puts(str[results.at(i)]);
}
int main(void) {
  scanf("%d", &n);
  int m = 0;
  for (int i = 0; i < n; i++) {
    char s[MAXL];
    scanf("%s", s);
    bool same = false;
    for (int j = 0; j < i; j++)
      if (!strcmp(s, str[j])) {
        ++times[j];
        same = true;
        break;
      }
    int l = strlen(s);
    if (!same) {
      memcpy(str[m], s, l * sizeof(char));
      str[m][l] = '\0';
      times[m++] = 1;
    }
  }
  n = m;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      if (i != j && check(i, j)) {
        v[i].push_back(j);
        ++deg[j];
      }
  out();
  return 0;
}

### 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;
const int MAXN = 100 + 10;
const int MAXL = 100 + 10;
char str[MAXN][MAXL];
vector<int> v[MAXN], results;
int n, times[MAXN], deg[MAXN];
bool check(int x, int y) {
  int n = strlen(str[x]), m = strlen(str[y]);
  for (int i = 0; i + n <= m; i++) {
    bool ok = true;
    for (int j = 0; j < n && ok; j++)
      if (str[x][j] != str[y][i + j]) ok = false;
    if (ok) return true;
  }
  return false;
}
void out() {
  results.clear();
  queue<int> q;
  for (; !q.empty(); q.pop())
    ;
  for (int i = 0; i < n; i++)
    if (!deg[i]) {
      if (!q.empty()) {
        puts("NO");
        return;
      }
      q.push(i);
    }
  for (; !q.empty();) {
    int u = q.front();
    q.pop();
    results.push_back(u);
    for (int i = 0; i < (signed)v[u].size(); i++) {
      if (!(--deg[v[u].at(i)])) {
        if (!q.empty()) {
          puts("NO");
          return;
        }
        q.push(v[u].at(i));
      }
    }
  }
  puts("YES");
  for (int i = 0; i < (signed)results.size(); i++)
    for (int j = 0; j < times[results.at(i)]; j++) puts(str[results.at(i)]);
}
int main(void) {
  scanf("%d", &n);
  int m = 0;
  for (int i = 0; i < n; i++) {
    char s[MAXL];
    scanf("%s", s);
    bool same = false;
    for (int j = 0; j < i; j++)
      if (!strcmp(s, str[j])) {
        ++times[j];
        same = true;
        break;
      }
    int l = strlen(s);
    if (!same) {
      memcpy(str[m], s, l * sizeof(char));
      str[m][l] = '\0';
      times[m++] = 1;
    }
  }
  n = m;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      if (i != j && check(i, j)) {
        v[i].push_back(j);
        ++deg[j];
      }
  out();
  return 0;
}

```