#include <bits/stdc++.h>
using namespace std;
const int MAXA = 2 * 1000056;
const int MAXN = 1123;
int a[MAXN];
bool sieve[MAXA];
void makeSieve() {
  int i;
  memset(sieve, true, sizeof(sieve));
  sieve[0] = sieve[1] = false;
  int s = sqrt(MAXA);
  for (i = 2; i <= s; i++)
    if (sieve[i]) {
      for (int j = i * i; j < MAXA; j += i) {
        sieve[j] = false;
      }
    }
}
int main() {
  int n, i, cnt, ans, b, ansi, j;
  scanf("%d", &n);
  for (i = 0; i <= n - 1; i++) scanf("%d", &a[i]);
  makeSieve();
  cnt = 0;
  for (i = 0; i <= n - 1; i++)
    if (a[i] == 1) cnt++;
  ans = 0;
  if (cnt > 0) {
    for (i = 0; i <= n - 1; i++)
      if (!((a[i]) & 1)) {
        b = a[i] + 1;
        if (sieve[b]) {
          ans = cnt + 1;
          ansi = a[i];
          break;
        }
      }
    if (ans == 0 && cnt > 1) ans = cnt;
  }
  for (i = 0; i <= n - 2; i++)
    for (j = i + 1; j <= n - 1; j++) {
      b = a[i] + a[j];
      if (sieve[b]) {
        if (2 > ans) {
          printf("2\n");
          printf("%d ", a[i]);
          printf("%d\n", a[j]);
          return 0;
        }
      }
    }
  for (i = 0; i <= n - 1; i++)
    if (sieve[a[i]]) {
      if (1 > ans) {
        printf("1\n");
        printf("%d\n", a[i]);
        return 0;
      }
    }
  if (ans != 0) {
    printf("%d\n", ans);
    if (ans > 0) {
      for (i = 1; i <= cnt; i++) printf("1 ");
      if (ans > cnt)
        printf("%d\n", ansi);
      else
        printf("\n");
    }
  } else {
    ans = 1;
    printf("%d\n", ans);
    printf("%d\n", a[0]);
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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 MAXA = 2 * 1000056;
const int MAXN = 1123;
int a[MAXN];
bool sieve[MAXA];
void makeSieve() {
  int i;
  memset(sieve, true, sizeof(sieve));
  sieve[0] = sieve[1] = false;
  int s = sqrt(MAXA);
  for (i = 2; i <= s; i++)
    if (sieve[i]) {
      for (int j = i * i; j < MAXA; j += i) {
        sieve[j] = false;
      }
    }
}
int main() {
  int n, i, cnt, ans, b, ansi, j;
  scanf("%d", &n);
  for (i = 0; i <= n - 1; i++) scanf("%d", &a[i]);
  makeSieve();
  cnt = 0;
  for (i = 0; i <= n - 1; i++)
    if (a[i] == 1) cnt++;
  ans = 0;
  if (cnt > 0) {
    for (i = 0; i <= n - 1; i++)
      if (!((a[i]) & 1)) {
        b = a[i] + 1;
        if (sieve[b]) {
          ans = cnt + 1;
          ansi = a[i];
          break;
        }
      }
    if (ans == 0 && cnt > 1) ans = cnt;
  }
  for (i = 0; i <= n - 2; i++)
    for (j = i + 1; j <= n - 1; j++) {
      b = a[i] + a[j];
      if (sieve[b]) {
        if (2 > ans) {
          printf("2\n");
          printf("%d ", a[i]);
          printf("%d\n", a[j]);
          return 0;
        }
      }
    }
  for (i = 0; i <= n - 1; i++)
    if (sieve[a[i]]) {
      if (1 > ans) {
        printf("1\n");
        printf("%d\n", a[i]);
        return 0;
      }
    }
  if (ans != 0) {
    printf("%d\n", ans);
    if (ans > 0) {
      for (i = 1; i <= cnt; i++) printf("1 ");
      if (ans > cnt)
        printf("%d\n", ansi);
      else
        printf("\n");
    }
  } else {
    ans = 1;
    printf("%d\n", ans);
    printf("%d\n", a[0]);
  }
  return 0;
}

```