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;
vector<int> a[205];
int f[205][205], c[205][205], n, m, Trace[205];
bool bfs() {
  queue<int> qu;
  qu.push(n + 1);
  memset(Trace, 0, sizeof Trace);
  Trace[n + 1] = n + 1;
  while (qu.size()) {
    int u = qu.front();
    qu.pop();
    for (int i = 0; i < a[u].size(); ++i) {
      int v = a[u][i];
      if (Trace[v]) continue;
      if (f[u][v] < c[u][v]) qu.push(v), Trace[v] = u;
    }
  }
  int pos = n + 2;
  if (!Trace[pos]) return 0;
  int add = 1e9;
  while (Trace[pos] != pos)
    add = min(add, c[Trace[pos]][pos] - f[Trace[pos]][pos]), pos = Trace[pos];
  pos = n + 2;
  while (Trace[pos] != pos)
    f[Trace[pos]][pos] += add, f[pos][Trace[pos]] -= add, pos = Trace[pos];
  return true;
}
int p[205], cnt;
bool visited[205];
vector<int> ans[205];
int val[205];
bool check(int u) {
  int can = sqrt(u);
  for (int i = 2; i <= can + 1; ++i)
    if (u % i == 0) return false;
  return true;
}
void dfs(int u) {
  ans[cnt].push_back(u);
  visited[u] = true;
  for (int i = 0; i < a[u].size(); ++i)
    if (f[u][a[u][i]]) {
      if (visited[a[u][i]] || a[u][i] > n) continue;
      dfs(a[u][i]);
    }
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> val[i];
  for (int i = 1; i <= n; ++i)
    for (int j = i + 1; j <= n; ++j)
      if (check(val[i] + val[j])) {
        if (val[i] & 1)
          c[i][j] = 1;
        else
          c[j][i] = 1;
        a[i].push_back(j), a[j].push_back(i);
      }
  for (int i = 1; i <= n; ++i) {
    if (val[i] & 1)
      a[n + 1].push_back(i), a[i].push_back(n + 1), c[n + 1][i] = 2;
    else
      a[i].push_back(n + 2), a[n + 2].push_back(i), c[i][n + 2] = 2;
  }
  while (bfs())
    ;
  for (int i = 1; i <= n; ++i)
    if (val[i] % 2 == 0 && f[i][n + 2] != 2) return cout << "Impossible", 0;
  for (int i = 1; i <= n; ++i) {
    if (!(val[i] & 1)) continue;
    if (!visited[i]) cnt++, dfs(i);
  }
  cout << cnt << endl;
  for (int i = 1; i <= cnt; ++i) {
    cout << ans[i].size() << ' ';
    for (int j = 0; j < ans[i].size(); ++j) cout << ans[i][j] << ' ';
    cout << endl;
  }
}