#include <bits/stdc++.h>
using namespace std;
const int maxn = 1005;
int pri[2000005];
bool flag[maxn], a[maxn][maxn];
int ans, cnt[maxn], group[maxn], n, vis[maxn];
bool dfs(int u, int pos) {
  int i, j;
  for (i = u + 1; i <= n; i++) {
    if (cnt[i] + pos <= ans) return 0;
    if (a[u][i]) {
      for (j = 0; j < pos; j++)
        if (!a[i][vis[j]]) break;
      if (j == pos) {
        vis[pos] = i;
        if (dfs(i, pos + 1)) return 1;
      }
    }
  }
  if (pos > ans) {
    for (i = 0; i < pos; i++) group[i] = vis[i];
    ans = pos;
    return 1;
  }
  return 0;
}
void maxclique() {
  ans = -1;
  for (int i = n; i > 0; i--) {
    vis[0] = i;
    dfs(i, 1);
    cnt[i] = ans;
  }
}
void pre() {
  pri[1] = 1;
  pri[0] = 1;
  for (int i = 2; i < 2000005; i++) {
    if (pri[i]) continue;
    for (int j = i + i; j < 2000005; j += i) pri[j] = 1;
  }
}
int aa[maxn];
int main() {
  pre();
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &aa[i]);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j < i; j++)
      if (!pri[aa[i] + aa[j]]) a[i][j] = 1, a[j][i] = 1;
  maxclique();
  cout << ans << endl;
  for (int i = 0; i < ans; i++) cout << aa[group[i]] << " ";
  cout << endl;
}

### Prompt

Your challenge is to write 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 = 1005;
int pri[2000005];
bool flag[maxn], a[maxn][maxn];
int ans, cnt[maxn], group[maxn], n, vis[maxn];
bool dfs(int u, int pos) {
  int i, j;
  for (i = u + 1; i <= n; i++) {
    if (cnt[i] + pos <= ans) return 0;
    if (a[u][i]) {
      for (j = 0; j < pos; j++)
        if (!a[i][vis[j]]) break;
      if (j == pos) {
        vis[pos] = i;
        if (dfs(i, pos + 1)) return 1;
      }
    }
  }
  if (pos > ans) {
    for (i = 0; i < pos; i++) group[i] = vis[i];
    ans = pos;
    return 1;
  }
  return 0;
}
void maxclique() {
  ans = -1;
  for (int i = n; i > 0; i--) {
    vis[0] = i;
    dfs(i, 1);
    cnt[i] = ans;
  }
}
void pre() {
  pri[1] = 1;
  pri[0] = 1;
  for (int i = 2; i < 2000005; i++) {
    if (pri[i]) continue;
    for (int j = i + i; j < 2000005; j += i) pri[j] = 1;
  }
}
int aa[maxn];
int main() {
  pre();
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &aa[i]);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j < i; j++)
      if (!pri[aa[i] + aa[j]]) a[i][j] = 1, a[j][i] = 1;
  maxclique();
  cout << ans << endl;
  for (int i = 0; i < ans; i++) cout << aa[group[i]] << " ";
  cout << endl;
}

```