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 = 209, INF = 1e6;
int n, s, t, a[MAX], cf[MAX][MAX], f[MAX][MAX], o, e;
vector<int> ans;
bool pr(int x) {
  if (x == 1) return false;
  for (int i = 2; i * i <= x; i++)
    if (x % i == 0) return false;
  return true;
}
queue<int> q;
int fa[MAX], fl[MAX];
bool mark[MAX];
bool bfs() {
  memset(fa, -1, sizeof fa), fa[s] = s, fl[s] = INF, q.push(s);
  while (q.size()) {
    int v = q.front();
    q.pop();
    for (int i = 0; i < n + 2; i++)
      if (cf[v][i] > 0 && fa[i] == -1)
        fa[i] = v, q.push(i), fl[i] = min(cf[v][i], fl[v]);
  }
  if (fa[t] == -1) return false;
  int p = t;
  while (p != s)
    cf[fa[p]][p] -= fl[t], cf[p][fa[p]] += fl[t], f[fa[p]][p] += fl[t],
        f[p][fa[p]] -= fl[t], p = fa[p];
  return true;
}
int max_flow() {
  int ret = 0;
  while (bfs())
    ;
  for (int i = 0; i < n + 2; i++) ret += f[i][t];
  return ret;
}
void dfs2(int v) {
  mark[v] = true;
  ans.push_back(v);
  for (int i = 0; i < n; i++)
    if ((f[v][i] || f[i][v]) && !mark[i]) dfs2(i);
}
stringstream ss;
int main() {
  cin >> n, s = n, t = n + 1;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] % 2 == 1)
      cf[s][i] = 2, o++;
    else
      cf[i][t] = 2, e++;
  }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      if (a[i] % 2 == 1 && a[j] % 2 == 0 && pr(a[i] + a[j])) cf[i][j] = 1;
  int mx = max_flow();
  if (mx != n || o != e) return cout << "Impossible\n", 0;
  memset(mark, 0, sizeof mark);
  int l = 0;
  for (int i = 0; i < n; i++)
    if (!mark[i]) {
      l++;
      ans.clear(), dfs2(i), ss << ans.size() << ' ';
      for (auto i : ans) ss << i + 1 << ' ';
      ss << '\n';
    }
  cout << l << '\n';
  for (int i = 0; i < l; i++) {
    int x;
    ss >> x, cout << x << ' ';
    for (int i = 0, v; i < x; i++) ss >> v, cout << v << ' ';
    cout << '\n';
  }
}