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;
bool p[300005], vis[300005];
int n, a[405], c[405][405], source, sink, f[405][405], d[405], num;
vector<int> ans[405];
vector<int> graph[405];
queue<int> q;
bool bfs() {
  for (int i = 1; i <= 2 * n + 2; i++) d[i] = 0;
  d[source] = 1;
  q.push(source);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int i = 0; i < graph[u].size(); i++) {
      int v = graph[u][i];
      if (!d[v] && f[u][v] < c[u][v]) {
        d[v] = u;
        q.push(v);
      }
    }
  }
  return (d[sink]);
}
void augment() {
  int u = sink, v, a = 1e9;
  while (u != source) {
    v = d[u];
    a = min(a, c[v][u] - f[v][u]);
    u = v;
  }
  u = sink;
  while (u != source) {
    v = d[u];
    f[v][u] += a;
    f[u][v] -= a;
    u = v;
  }
}
void dfs(int u) {
  vis[u] = 1;
  ans[num].push_back(u);
  for (int i = 0; i < graph[u].size(); i++) {
    int v = graph[u][i];
    if (!vis[v] && abs(f[u][v]) == 1) dfs(v);
  }
}
int main() {
  cin.tie(0), ios::sync_with_stdio(0);
  for (int i = 2; i <= 20000; i++) p[i] = 1;
  for (int i = 2; i <= 20000; i++) {
    if (p[i]) {
      for (int j = 2 * i; j <= 20000; j += i) p[j] = 0;
    }
  }
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  source = 2 * n + 1;
  sink = 2 * n + 2;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (a[i] % 2 == 0 && p[a[i] + a[j]]) {
        c[i][j] = 1;
        graph[i].push_back(j);
        graph[j].push_back(i);
      }
    }
    if (a[i] % 2 == 0 && graph[i].size() < 2) {
      cout << "Impossible\n";
      exit(0);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (a[i] % 2 == 0) {
      graph[source].push_back(i);
      graph[i].push_back(source);
      c[source][i] = 2;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (a[i] % 2 != 0) {
      graph[sink].push_back(i);
      graph[i].push_back(sink);
      c[i][sink] = 2;
    }
  }
  while (bfs()) augment();
  int sum = 0;
  for (int i = 0; i < graph[source].size(); i++) {
    sum += f[source][graph[source][i]];
    if (f[source][graph[source][i]] != 2) {
      cout << "Impossible\n";
      exit(0);
    }
  }
  if (sum != n) {
    cout << "Impossible\n";
    exit(0);
  }
  for (int i = 1; i <= n; i++) {
    if (a[i] % 2 == 0 && vis[i] == 0) {
      dfs(i);
      num++;
    }
  }
  cout << num << "\n";
  for (int i = 0; i < num; i++) {
    cout << ans[i].size() << " ";
    for (int j = 0; j < ans[i].size(); j++) cout << ans[i][j] << " ";
    cout << "\n";
  }
}