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 MAX_N = 505;
const int MAX_V = 25000;
const int INF = 1000000000;
const int SRC = MAX_N - 1;
const int DST = MAX_N - 2;
bool vis[MAX_N], prime[MAX_V];
int n, a[MAX_N], cap[MAX_N][MAX_N], flow[MAX_N][MAX_N];
vector<int> e[MAX_N];
int dfs(int u, int bottle) {
  if (u == DST) {
    return bottle;
  }
  vis[u] = true;
  for (int i = 0; i < MAX_N; i++) {
    if (!vis[i] && flow[u][i] < cap[u][i]) {
      int tmp = dfs(i, min(bottle, cap[u][i] - flow[u][i]));
      if (tmp > 0) {
        flow[u][i] += tmp;
        flow[i][u] -= tmp;
        return tmp;
      }
    }
  }
  return 0;
}
void find_cycle(int u, vector<int> &cycle) {
  vis[u] = true;
  cycle.push_back(u);
  for (int i = 0; i < e[u].size(); i++) {
    if (!vis[e[u][i]]) {
      find_cycle(e[u][i], cycle);
    }
  }
}
int main() {
  memset(prime, true, sizeof(prime));
  prime[0] = prime[1] = false;
  for (int i = 2; i < MAX_V; i++) {
    if (!prime[i]) {
      continue;
    }
    for (int j = i * i; j < MAX_V; j += i) {
      prime[j] = false;
    }
  }
  cin >> n;
  int num_o = 0, num_e = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] % 2 == 0) {
      num_e++;
      cap[SRC][i] = 2;
    } else {
      num_o++;
      cap[n + i][DST] = 2;
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (a[i] % 2 == 0 && a[j] % 2 == 1 && prime[a[i] + a[j]]) {
        cap[i][n + j] = 1;
      }
    }
  }
  int fl = 0;
  while (1) {
    memset(vis, false, sizeof(vis));
    int tmp = dfs(SRC, INF);
    if (tmp == 0) {
      break;
    }
    fl += tmp;
  }
  if (num_o != num_e || fl != num_o * 2) {
    cout << "Impossible" << endl;
  } else {
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        if (flow[i][n + j] == 1) {
          e[i].push_back(j);
          e[j].push_back(i);
        }
      }
    }
    memset(vis, false, sizeof(vis));
    vector<vector<int> > cycles;
    for (int i = 0; i < n; i++) {
      if (!vis[i]) {
        vector<int> cycle;
        find_cycle(i, cycle);
        cycles.push_back(cycle);
      }
    }
    cout << cycles.size() << endl;
    for (int i = 0; i < cycles.size(); i++) {
      cout << cycles[i].size() << " ";
      for (int j = 0; j < cycles[i].size(); j++) {
        cout << cycles[i][j] + 1 << " ";
      }
      cout << endl;
    }
  }
  return 0;
}