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;
const int M = 200 + 4, Inf = 1e9 + 10;
const int N = M;
int viz[N], mat[N][N], network[N][N], parent[N];
bool anotherPath(int start, int end) {
  memset(viz, 0, sizeof viz);
  memset(parent, -1, sizeof parent);
  viz[start] = true;
  queue<int> q;
  q.push(start);
  while (!q.empty()) {
    int z = q.front();
    q.pop();
    viz[z] = true;
    for (int i = 0; i < N; i++) {
      if (network[z][i] <= 0 || viz[i]) continue;
      viz[i] = true;
      parent[i] = z;
      if (i == end) return true;
      q.push(i);
    }
  }
  return false;
}
int maxflow(int start, int end) {
  memcpy(network, mat, sizeof(mat));
  int total = 0;
  while (anotherPath(start, end)) {
    int flow = network[parent[end]][end];
    int curr = end;
    while (parent[curr] >= 0) {
      flow = min(flow, network[parent[curr]][curr]);
      curr = parent[curr];
    }
    curr = end;
    while (parent[curr] >= 0) {
      network[parent[curr]][curr] -= flow;
      network[curr][parent[curr]] += flow;
      curr = parent[curr];
    }
    total += flow;
  }
  return total;
}
int a[M];
int n;
bool isprime(int x) {
  bool p = 1;
  for (int i = 2; p && i * i <= x; i++)
    if (!(x % i)) p = 0;
  return p;
}
bool mark[M];
vector<int> cyc;
vector<vector<int> > ans;
void dfs(int v) {
  mark[v] = 1;
  cyc.push_back(v);
  for (int i = 0; i < n; i++)
    if (!mark[i] &&
        ((mat[i][v] && !network[i][v]) || (mat[v][i] && !network[v][i]))) {
      dfs(i);
      return;
    }
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n && !(a[i] % 2); j++)
      if (isprime(a[i] + a[j])) mat[i][j] = 1;
  if (n % 2) {
    cout << "Impossible\n";
    return 0;
  }
  for (int i = 0; i < n; i++)
    if (!(a[i] % 2)) mat[n][i] = 2;
  for (int i = 0; i < n; i++)
    if (a[i] % 2) mat[i][n + 1] = 2;
  if (maxflow(n, n + 1) == n) {
    for (int i = 0; i < n; i++)
      if (!mark[i]) {
        cyc.clear();
        dfs(i);
        ans.push_back(cyc);
      }
    cout << ans.size() << endl;
    for (int i = 0; i < ans.size(); i++) {
      cout << ans[i].size() << " ";
      for (int j = 0; j < ans[i].size(); j++) cout << 1 + ans[i][j] << " ";
      cout << endl;
    }
  } else
    cout << "Impossible\n";
  return 0;
}
