Problem: Fox Ciel is participating in a party in Prime Kingdom. There are n foxes there (include Fox Ciel). The i-th fox is ai years old.

They will have dinner around some round tables. You want to distribute foxes such that:

  1. Each fox is sitting at some table. 
  2. Each table has at least 3 foxes sitting around it. 
  3. The sum of ages of any two adjacent foxes around each table should be a prime number. 



If k foxes f1, f2, ..., fk are sitting around table in clockwise order, then for 1 ≤ i ≤ k - 1: fi and fi + 1 are adjacent, and f1 and fk are also adjacent.

If it is possible to distribute the foxes in the desired manner, find out a way to do that.

Input

The first line contains single integer n (3 ≤ n ≤ 200): the number of foxes in this party. 

The second line contains n integers ai (2 ≤ ai ≤ 104).

Output

If it is impossible to do this, output "Impossible".

Otherwise, in the first line output an integer m (<image>): the number of tables.

Then output m lines, each line should start with an integer k -=– the number of foxes around that table, and then k numbers — indices of fox sitting around that table in clockwise order.

If there are several possible arrangements, output any of them.

Examples

Input

4
3 4 8 9


Output

1
4 1 2 4 3


Input

5
2 2 2 2 2


Output

Impossible


Input

12
2 3 4 5 6 7 8 9 10 11 12 13


Output

1
12 1 2 3 6 5 12 9 8 7 10 11 4


Input

24
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25


Output

3
6 1 2 3 6 5 4
10 7 8 9 12 15 14 13 16 11 10
8 17 18 23 22 19 20 21 24

Note

In example 1, they can sit around one table, their ages are: 3-8-9-4, adjacent sums are: 11, 17, 13 and 7, all those integers are primes.

In example 2, it is not possible: the sum of 2+2 = 4 is not a prime number.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 40004;
int i, j, lp[N], f[N], cnt, a[N], n, p[N];
int c[202][202], sz;
vector<int> pr, ans[202];
bool go(int u) {
  if (p[u]) return 0;
  p[u] = 1;
  if (u == n + 1) return 1;
  for (int i = 0; i <= n + 1; i++)
    if (c[u][i] > 0 && go(i)) {
      c[u][i]--;
      c[i][u]++;
      return 1;
    }
  return 0;
}
void dfs(int u) {
  p[u] = 1;
  ans[sz].push_back(u);
  for (int i = 1; i <= n; i++) {
    if (p[i]) continue;
    int x = u, y = i;
    if (a[y] & 1) swap(x, y);
    if (!f[a[x] + a[y]]) continue;
    if (c[x][y] < 1) {
      dfs(i);
      return;
    }
  }
}
int main() {
  n = 40000;
  for (i = 2; i < n; i++) {
    if (!lp[i]) lp[i] = i, pr.push_back(i);
    for (j = 0; j < pr.size() && pr[j] <= lp[i] && i * pr[j] <= n; j++)
      lp[i * pr[j]] = pr[j];
  }
  for (i = 0; i < pr.size(); i++) f[pr[i]] = 1;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (i = 1; i <= n; i++)
    if (a[i] & 1)
      for (j = 1; j <= n; j++)
        if (!(a[j] & 1) && f[a[j] + a[i]]) c[i][j] = 1;
  for (i = 1; i <= n; i++)
    if (a[i] & 1)
      c[0][i] = 2, cnt++;
    else
      c[i][n + 1] = 2;
  if (2 * cnt != n) {
    cout << "Impossible" << endl;
    return 0;
  }
  for (i = 1; i <= n; i++) {
    if (!go(0)) {
      cout << "Impossible" << endl;
      return 0;
    }
    for (j = 0; j <= n + 1; j++) p[j] = 0;
  }
  for (i = 1; i <= n; i++)
    if (!p[i]) {
      sz++;
      dfs(i);
    }
  printf("%d\n", sz);
  for (i = 1; i <= sz; i++) {
    cout << ans[i].size() << ' ';
    for (j = 0; j < ans[i].size(); j++) cout << ans[i][j] << ' ';
    cout << endl;
  }
  return 0;
}