#include <bits/stdc++.h>
int a[1000000];
int isntprime[3000001], tot_prime, prime[3000001];
void sift(int n) {
  for (int i = 2; i <= n; ++i) {
    if (!isntprime[i]) prime[++tot_prime] = i;
    for (int j = 1; j <= tot_prime && 1LL * prime[j] * i <= n; ++j) {
      isntprime[i * prime[j]] = 1;
      if (i % prime[j] == 0) break;
    }
  }
}
int main() {
  sift(2000001);
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  int cnt = 0;
  for (int i = 1; i <= n; ++i) cnt += a[i] == 1;
  if (cnt < 2) {
    int flag1 = 0, flag2 = 0, p1, p2;
    for (int i = 1; i <= n; ++i) {
      for (int j = i + 1; j <= n; ++j)
        if (!isntprime[a[i] + a[j]]) flag2 = 1, p1 = i, p2 = j;
    }
    if (flag2) {
      printf("2\n");
      printf("%d %d\n", a[p1], a[p2]);
    } else {
      printf("1\n%d\n", a[1]);
    }
  } else {
    int flag = 0;
    for (int j = 1; j <= n; ++j) {
      if (a[j] != 1 && !isntprime[a[j] + 1]) flag = j;
    }
    if (flag) {
      printf("%d\n", cnt + 1);
      for (int i = 1; i <= cnt; ++i) printf("%d ", 1);
      printf("%d\n", a[flag]);
    } else {
      printf("%d\n", cnt);
      for (int i = 1; i <= cnt; ++i) printf("%d ", 1);
      printf("\n");
    }
  }
  scanf("%d", &n);
}

### Prompt

Create a solution in Cpp for 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>
int a[1000000];
int isntprime[3000001], tot_prime, prime[3000001];
void sift(int n) {
  for (int i = 2; i <= n; ++i) {
    if (!isntprime[i]) prime[++tot_prime] = i;
    for (int j = 1; j <= tot_prime && 1LL * prime[j] * i <= n; ++j) {
      isntprime[i * prime[j]] = 1;
      if (i % prime[j] == 0) break;
    }
  }
}
int main() {
  sift(2000001);
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  int cnt = 0;
  for (int i = 1; i <= n; ++i) cnt += a[i] == 1;
  if (cnt < 2) {
    int flag1 = 0, flag2 = 0, p1, p2;
    for (int i = 1; i <= n; ++i) {
      for (int j = i + 1; j <= n; ++j)
        if (!isntprime[a[i] + a[j]]) flag2 = 1, p1 = i, p2 = j;
    }
    if (flag2) {
      printf("2\n");
      printf("%d %d\n", a[p1], a[p2]);
    } else {
      printf("1\n%d\n", a[1]);
    }
  } else {
    int flag = 0;
    for (int j = 1; j <= n; ++j) {
      if (a[j] != 1 && !isntprime[a[j] + 1]) flag = j;
    }
    if (flag) {
      printf("%d\n", cnt + 1);
      for (int i = 1; i <= cnt; ++i) printf("%d ", 1);
      printf("%d\n", a[flag]);
    } else {
      printf("%d\n", cnt);
      for (int i = 1; i <= cnt; ++i) printf("%d ", 1);
      printf("\n");
    }
  }
  scanf("%d", &n);
}

```