#include <bits/stdc++.h>
using namespace std;
void codeforces_486_A() {
  int n, k;
  scanf("%d %d", &n, &k);
  int a[n];
  vector<int> b;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    int j;
    for (j = 0; j < i; j++) {
      if (a[i] == a[j]) {
        break;
      }
    }
    if (j == i) {
      b.push_back(i + 1);
    }
  }
  if (b.size() < k) {
    printf("NO");
  } else {
    printf("YES\n");
    for (int i = 0; i < k; i++) {
      printf("%d ", b[i]);
    }
  }
}
struct pppair {
  char c[101];
  int len;
};
bool operator<(const pppair a, const pppair b) { return a.len < b.len; }
void codeforces_486_B() {
  int n;
  scanf("%d", &n);
  pppair p[n];
  for (int i = 0; i < n; i++) {
    scanf("%s", p[i].c);
    p[i].len = strlen(p[i].c);
  }
  sort(p, p + n);
  int i;
  for (i = 0; i < n - 1; i++) {
    int yes = 0;
    for (int j = 0; j <= p[i + 1].len - p[i].len; j++) {
      int no = 0;
      for (int k = 0; k < p[i].len; k++) {
        if (p[i].c[k] != p[i + 1].c[k + j]) {
          no = 1;
        }
      }
      if (no == 0) {
        yes = 1;
      }
    }
    if (yes == 0) {
      break;
    }
  }
  if (i == n - 1) {
    printf("YES\n");
    for (int ii = 0; ii < n; ii++) {
      printf("%s\n", p[ii].c);
    }
  } else {
    printf("NO");
  }
}
int main() {
  codeforces_486_B();
  return 0;
}

### Prompt

Your task is to create 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 namespace std;
void codeforces_486_A() {
  int n, k;
  scanf("%d %d", &n, &k);
  int a[n];
  vector<int> b;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    int j;
    for (j = 0; j < i; j++) {
      if (a[i] == a[j]) {
        break;
      }
    }
    if (j == i) {
      b.push_back(i + 1);
    }
  }
  if (b.size() < k) {
    printf("NO");
  } else {
    printf("YES\n");
    for (int i = 0; i < k; i++) {
      printf("%d ", b[i]);
    }
  }
}
struct pppair {
  char c[101];
  int len;
};
bool operator<(const pppair a, const pppair b) { return a.len < b.len; }
void codeforces_486_B() {
  int n;
  scanf("%d", &n);
  pppair p[n];
  for (int i = 0; i < n; i++) {
    scanf("%s", p[i].c);
    p[i].len = strlen(p[i].c);
  }
  sort(p, p + n);
  int i;
  for (i = 0; i < n - 1; i++) {
    int yes = 0;
    for (int j = 0; j <= p[i + 1].len - p[i].len; j++) {
      int no = 0;
      for (int k = 0; k < p[i].len; k++) {
        if (p[i].c[k] != p[i + 1].c[k + j]) {
          no = 1;
        }
      }
      if (no == 0) {
        yes = 1;
      }
    }
    if (yes == 0) {
      break;
    }
  }
  if (i == n - 1) {
    printf("YES\n");
    for (int ii = 0; ii < n; ii++) {
      printf("%s\n", p[ii].c);
    }
  } else {
    printf("NO");
  }
}
int main() {
  codeforces_486_B();
  return 0;
}

```