#include <bits/stdc++.h>
using namespace std;
const int maxn = 105;
string a[maxn];
char A[maxn];
char B[maxn];
int panduan(int one, int two) {
  int x = a[one].size();
  int y = a[two].size();
  int pa = 0, pb = 0;
  int key = 0;
  for (string::iterator it = a[one].begin(); it != a[one].end(); it++) {
    A[pa] = *it;
    pa++;
  }
  for (string::iterator iit = a[two].begin(); iit != a[two].end(); iit++) {
    B[pb] = *iit;
    pb++;
  }
  pa = 0;
  pb = 0;
  while (pa < x && pb < y) {
    if (A[pa] == B[pb]) {
      pa++;
      pb++;
    } else {
      key++;
      int c = 0;
      for (int i = 0; i < y; i++) {
        if (B[i] == A[0]) c++;
        if (c == key) {
          pb = i;
          break;
        }
      }
      if (c < key) pb++;
      pa = 0;
    }
  }
  if (pa == x)
    return 1;
  else
    return 0;
}
int main() {
  int n;
  while (cin >> n) {
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    int IF = 0;
    for (int i = 0; i < n - 1; i++) {
      for (int j = 0; j < n - 1 - i; j++) {
        string b = a[j];
        string c = a[j + 1];
        int x = b.size();
        int y = c.size();
        if (x > y) {
          swap(a[j], a[j + 1]);
        } else if (x == y) {
          if (!panduan(j, j + 1)) {
            IF = 1;
          }
        }
      }
    }
    if (IF == 0) {
      for (int i = 0; i < n - 1; i++) {
        if (!panduan(i, i + 1)) {
          IF = 1;
          break;
        }
      }
    }
    if (IF == 0) {
      cout << "YES" << endl;
      for (int i = 0; i < n; i++) {
        cout << a[i] << endl;
      }
    } else {
      cout << "NO" << endl;
    }
  }
  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 = 105;
string a[maxn];
char A[maxn];
char B[maxn];
int panduan(int one, int two) {
  int x = a[one].size();
  int y = a[two].size();
  int pa = 0, pb = 0;
  int key = 0;
  for (string::iterator it = a[one].begin(); it != a[one].end(); it++) {
    A[pa] = *it;
    pa++;
  }
  for (string::iterator iit = a[two].begin(); iit != a[two].end(); iit++) {
    B[pb] = *iit;
    pb++;
  }
  pa = 0;
  pb = 0;
  while (pa < x && pb < y) {
    if (A[pa] == B[pb]) {
      pa++;
      pb++;
    } else {
      key++;
      int c = 0;
      for (int i = 0; i < y; i++) {
        if (B[i] == A[0]) c++;
        if (c == key) {
          pb = i;
          break;
        }
      }
      if (c < key) pb++;
      pa = 0;
    }
  }
  if (pa == x)
    return 1;
  else
    return 0;
}
int main() {
  int n;
  while (cin >> n) {
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    int IF = 0;
    for (int i = 0; i < n - 1; i++) {
      for (int j = 0; j < n - 1 - i; j++) {
        string b = a[j];
        string c = a[j + 1];
        int x = b.size();
        int y = c.size();
        if (x > y) {
          swap(a[j], a[j + 1]);
        } else if (x == y) {
          if (!panduan(j, j + 1)) {
            IF = 1;
          }
        }
      }
    }
    if (IF == 0) {
      for (int i = 0; i < n - 1; i++) {
        if (!panduan(i, i + 1)) {
          IF = 1;
          break;
        }
      }
    }
    if (IF == 0) {
      cout << "YES" << endl;
      for (int i = 0; i < n; i++) {
        cout << a[i] << endl;
      }
    } else {
      cout << "NO" << endl;
    }
  }
  return 0;
}

```