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;
int n;
bool pr[100100];
int a[100100];
vector<int> p[100100];
struct ed {
  int first, to, ca, fl;
};
int k;
ed edg[100100];
int dist[100100];
bool bfs() {
  queue<int> q;
  for (int i = 0; i <= n + 1; i++) {
    dist[i] = 1e9;
  }
  dist[0] = 0;
  q.push(0);
  while (!q.empty()) {
    int t = q.front();
    q.pop();
    for (int i = 0; i < p[t].size(); i++) {
      int in = p[t][i];
      if (dist[edg[in].to] > dist[t] + 1 && edg[in].fl < edg[in].ca) {
        dist[edg[in].to] = dist[t] + 1;
        q.push(edg[in].to);
      }
    }
  }
  return dist[n + 1] < 1e8;
}
int it[100100];
int dfs(int x, int fl) {
  if (!fl) return 0;
  if (x == n + 1) return fl;
  for (; it[x] < p[x].size(); it[x]++) {
    int in = p[x][it[x]];
    if (dist[edg[in].to] != dist[x] + 1) continue;
    int t = dfs(edg[in].to, min(fl, edg[in].ca - edg[in].fl));
    if (t) {
      edg[in].fl += t;
      edg[in ^ 1].fl -= t;
      return t;
    }
  }
  return 0;
}
int maxflow() {
  int r = 0;
  while (bfs()) {
    memset(it, 0, sizeof it);
    int y;
    while (y = dfs(0, 1e9)) r += y;
  }
  return r;
}
bool v[100100];
vector<vector<int> > res;
void print(int x) {
  if (x == 0) return;
  if (x == n + 1) return;
  if (v[x]) return;
  v[x] = 1;
  res.back().push_back(x);
  for (int i = 0; i < p[x].size(); i++) {
    if (edg[p[x][i]].fl) {
      print(edg[p[x][i]].to);
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  for (int i = 2; i < 100100; i++) {
    if (!pr[i]) {
      for (int j = i * 2; j < 100100; j += i) pr[j] = 1;
    }
  }
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      if (pr[a[i] + a[j]]) continue;
      if (a[i] % 2 != 0) {
        p[i].push_back(k);
        edg[k].first = i;
        edg[k].to = j;
        edg[k].ca = 1;
        edg[k].fl = 0;
        k++;
        p[j].push_back(k);
        edg[k].first = j;
        edg[k].to = i;
        edg[k].ca = 0;
        edg[k].fl = 0;
        k++;
      } else {
        p[j].push_back(k);
        edg[k].first = j;
        edg[k].to = i;
        edg[k].ca = 1;
        edg[k].fl = 0;
        k++;
        p[i].push_back(k);
        edg[k].first = i;
        edg[k].to = j;
        edg[k].ca = 0;
        edg[k].fl = 0;
        k++;
      }
    }
  }
  int d = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] % 2 != 0) {
      d++;
      p[0].push_back(k);
      edg[k].first = 0;
      edg[k].to = i;
      edg[k].ca = 2;
      edg[k].fl = 0;
      k++;
      p[i].push_back(k);
      edg[k].first = i;
      edg[k].to = 0;
      edg[k].ca = 0;
      edg[k].fl = 0;
      k++;
    } else {
      p[i].push_back(k);
      edg[k].first = i;
      edg[k].to = n + 1;
      edg[k].ca = 2;
      edg[k].fl = 0;
      k++;
      p[n + 1].push_back(k);
      edg[k].first = n + 1;
      edg[k].to = i;
      edg[k].ca = 0;
      edg[k].fl = 0;
      k++;
    }
  }
  int f = maxflow();
  if (f != n || d != n / 2) {
    cout << "Impossible" << endl;
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    if (!v[i]) {
      res.push_back(vector<int>());
      print(i);
    }
  }
  cout << res.size() << endl;
  for (int i = 0; i < res.size(); i++) {
    cout << res[i].size();
    for (int j = 0; j < res[i].size(); j++) {
      cout << " " << res[i][j];
    }
    cout << "\n";
  }
  return 0;
}