#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e6, M = 2e6;
int a[maxn], b[maxn], ans[maxn];
int la, i, j, e, bl, n, x;
int main() {
  b[0] = b[1] = 1;
  for (i = 2; i <= M; i++) {
    if (b[i]) continue;
    long long p = (long long)i * i;
    while (p <= M) {
      b[p] = 1;
      p += i;
    }
  }
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 1) x++;
  }
  if (n == 1) {
    printf("1\n%d", a[1]);
    return 0;
  }
  for (i = 1; i <= n; i++)
    for (j = i + 1; j <= n; j++) {
      if (b[a[i] + a[j]]) continue;
      ans[++la] = a[i];
      ans[++la] = a[j];
      goto pn;
    }
pn:
  if (x) {
    for (i = 1; i <= n; i++) {
      if (a[i] == 1) continue;
      if (b[a[i] + 1]) continue;
      bl = a[i];
    }
    if (bl) {
      printf("%d\n", x + 1);
      printf("%d", bl);
      for (i = 1; i <= x; i++) printf(" %d", 1);
    } else if (x > 1) {
      printf("%d\n", x);
      for (i = 1; i <= x; i++) printf("%d%c", 1, i < x ? ' ' : '\n');
    } else if (la) {
      printf("2\n");
      printf("%d %d\n", ans[1], ans[2]);
    } else
      printf("1\n%d", a[1]);
  } else {
    if (la) {
      printf("%d\n", la);
      printf("%d %d", ans[1], ans[2]);
    } else {
      printf("1\n");
      printf("%d", a[1]);
    }
  }
  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>
using namespace std;
const int maxn = 3e6, M = 2e6;
int a[maxn], b[maxn], ans[maxn];
int la, i, j, e, bl, n, x;
int main() {
  b[0] = b[1] = 1;
  for (i = 2; i <= M; i++) {
    if (b[i]) continue;
    long long p = (long long)i * i;
    while (p <= M) {
      b[p] = 1;
      p += i;
    }
  }
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 1) x++;
  }
  if (n == 1) {
    printf("1\n%d", a[1]);
    return 0;
  }
  for (i = 1; i <= n; i++)
    for (j = i + 1; j <= n; j++) {
      if (b[a[i] + a[j]]) continue;
      ans[++la] = a[i];
      ans[++la] = a[j];
      goto pn;
    }
pn:
  if (x) {
    for (i = 1; i <= n; i++) {
      if (a[i] == 1) continue;
      if (b[a[i] + 1]) continue;
      bl = a[i];
    }
    if (bl) {
      printf("%d\n", x + 1);
      printf("%d", bl);
      for (i = 1; i <= x; i++) printf(" %d", 1);
    } else if (x > 1) {
      printf("%d\n", x);
      for (i = 1; i <= x; i++) printf("%d%c", 1, i < x ? ' ' : '\n');
    } else if (la) {
      printf("2\n");
      printf("%d %d\n", ans[1], ans[2]);
    } else
      printf("1\n%d", a[1]);
  } else {
    if (la) {
      printf("%d\n", la);
      printf("%d %d", ans[1], ans[2]);
    } else {
      printf("1\n");
      printf("%d", a[1]);
    }
  }
  return 0;
}

```