#include <bits/stdc++.h>
char composite[2 * 1000000 + 1];
int main() {
  int m, n, *a, i, j, b, cnt = 0;
  for (n = 2; n <= 2 * 1000000; n++) {
    if (composite[n]) continue;
    for (m = n + n; m <= 2 * 1000000; m += n) composite[m] = 1;
  }
  scanf("%d", &n);
  a = (int *)malloc(sizeof(int) * n);
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 1) cnt++;
  }
  if (cnt > 1) {
    b = 0;
    for (i = 0; i < n; i++)
      if (a[i] > 1 && !composite[a[i] + 1]) b = a[i];
    printf("%d\n", b == 0 ? cnt : cnt + 1);
    printf("1");
    for (j = 1; j < cnt; j++) printf(" 1");
    if (b > 0) printf(" %d", b);
    printf("\n");
    return 0;
  }
  for (i = 0; i < n; i++)
    for (j = i + 1; j < n; j++)
      if (!composite[a[i] + a[j]]) {
        printf("2\n");
        printf("%d %d\n", a[i], a[j]);
        return 0;
      }
  printf("1\n");
  printf("%d\n", a[0]);
  return 0;
}

### Prompt

Generate a CPP solution to the following problem:

A tuple of positive integers {x1, x2, ..., xk} is called simple if for all pairs of positive integers (i, j) (1 ≤ i < j ≤ k), xi + xj is a prime.

You are given an array a with n positive integers a1, a2, ..., an (not necessary distinct). You want to find a simple subset of the array a with the maximum size.

A prime number (or a prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself.

Let's define a subset of the array a as a tuple that can be obtained from a by removing some (possibly all) elements of it.

Input

The first line contains integer n (1 ≤ n ≤ 1000) — the number of integers in the array a.

The second line contains n integers ai (1 ≤ ai ≤ 106) — the elements of the array a.

Output

On the first line print integer m — the maximum possible size of simple subset of a.

On the second line print m integers bl — the elements of the simple subset of the array a with the maximum size.

If there is more than one solution you can print any of them. You can print the elements of the subset in any order.

Examples

Input

2
2 3


Output

2
3 2


Input

2
2 2


Output

1
2


Input

3
2 1 1


Output

3
1 1 2


Input

2
83 14


Output

2
14 83

### Response

```cpp
#include <bits/stdc++.h>
char composite[2 * 1000000 + 1];
int main() {
  int m, n, *a, i, j, b, cnt = 0;
  for (n = 2; n <= 2 * 1000000; n++) {
    if (composite[n]) continue;
    for (m = n + n; m <= 2 * 1000000; m += n) composite[m] = 1;
  }
  scanf("%d", &n);
  a = (int *)malloc(sizeof(int) * n);
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 1) cnt++;
  }
  if (cnt > 1) {
    b = 0;
    for (i = 0; i < n; i++)
      if (a[i] > 1 && !composite[a[i] + 1]) b = a[i];
    printf("%d\n", b == 0 ? cnt : cnt + 1);
    printf("1");
    for (j = 1; j < cnt; j++) printf(" 1");
    if (b > 0) printf(" %d", b);
    printf("\n");
    return 0;
  }
  for (i = 0; i < n; i++)
    for (j = i + 1; j < n; j++)
      if (!composite[a[i] + a[j]]) {
        printf("2\n");
        printf("%d %d\n", a[i], a[j]);
        return 0;
      }
  printf("1\n");
  printf("%d\n", a[0]);
  return 0;
}

```