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.
#include <bits/stdc++.h>
using namespace std;
int n, f[444][444], c[444][444], a[200], s, t, even = 0, odd = 0;
int cnt = 0;
vector<vector<int> > ans;
vector<vector<int> > g(222);
char visit[222];
void fail() {
  cout << "Impossible\n";
  exit(0);
}
char dfs(int v, int flow = 2) {
  if (flow == 0) return false;
  if (v == t) return true;
  if (visit[v]) return false;
  visit[v] = true;
  for (int to = 0; to <= n + 1; ++to) {
    char pushed = dfs(to, min(flow, c[v][to] - f[v][to]));
    if (pushed) {
      f[v][to]++;
      f[to][v]--;
      return pushed;
    }
  }
  return false;
}
char check(int a) {
  for (int i = 2; i * i <= a; ++i)
    if (a % i == 0) return false;
  return true;
}
int main() {
  cin >> n;
  s = n;
  t = n + 1;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] % 2) {
      c[i][t] = 2;
      ++odd;
      continue;
    }
    c[s][i] = 2;
    ++even;
  }
  for (int i = 0; i < n; i++) {
    if (a[i] % 2) continue;
    for (int j = 0; j < n; j++) {
      if (check(a[i] + a[j])) {
        c[i][j] = 1;
      }
    }
  }
  if (even != odd) fail();
  int mxflow = 0;
  while (dfs(s)) {
    ++mxflow;
    for (int i = 0; i <= n + 1; i++) visit[i] = false;
  }
  if (mxflow != n) fail();
  for (int i = 0; i < n; i++) {
    if (visit[i]) continue;
    vector<int> cur;
    cur.push_back(i);
    for (int j = 0; j < cur.size(); ++j) {
      visit[cur[j]] = true;
      int v = cur[j];
      for (int i = 0; i < n; i++) {
        if (f[v][i] && !visit[i]) {
          cur.push_back(i);
          break;
        }
      }
    }
    if (cur.size() < 3) fail();
    ans.push_back(cur);
  }
  cout << ans.size() << '\n';
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i].size() << ' ';
    for (int j = 0; j < ans[i].size(); j++) cout << ans[i][j] + 1 << ' ';
  }
  return 0;
}
