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 mxn = 222;
int a[mxn], dist[mxn][mxn], used[mxn], r[mxn][mxn], ring[mxn][mxn], cntgs[mxn];
int n, cntring, s, t;
inline bool isp(int x) {
  for (int i = 2; i * i <= x; ++i)
    if (x % i == 0) return 0;
  return 1;
}
inline bool find(int x) {
  if (x == t) return 1;
  if (used[x]) return 0;
  used[x] = 1;
  for (int i = 1; i <= t; ++i)
    if (dist[x][i] and find(i)) {
      --dist[x][i], ++dist[i][x];
      return 1;
    }
  return 0;
}
inline void dfs(int x) {
  if (used[x]) return;
  used[x] = 1, ring[cntring][cntgs[cntring]++] = x;
  for (int i = 1; i <= 200; ++i)
    if (r[x][i]) dfs(i);
}
int main() {
  cin >> n;
  s = 0, t = 201;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    if (a[i] & 1)
      dist[s][i] = 2;
    else
      dist[i][t] = 2;
  }
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= n; ++j)
      if (a[i] & 1 and !(a[j] % 2) and isp(a[i] + a[j])) dist[i][j] = 1;
  int tms = 0;
  for (; find(s);) ++tms, memset(used, 0, sizeof(used));
  if (tms < n) return puts("Impossible") * 0;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= n; ++j)
      if (a[i] & 1 and !(a[j] & 1) and isp(a[i] + a[j]) and !dist[i][j])
        r[i][j] = r[j][i] = 1;
  memset(used, 0, sizeof(used));
  for (int i = 1; i <= n; ++i)
    if (!used[i]) dfs(i), ++cntring;
  cout << cntring << endl;
  for (int i = 0; i < cntring; ++i) {
    cout << cntgs[i] << ' ';
    for (int j = 0; j < cntgs[i]; ++j) cout << ring[i][j] << ' ';
    cout << '\n';
  }
}
