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 N = 300;
const int oo = 1000;
int n, MaxFlow, cnt;
int pa[N];
vector<int> ed2[N];
int c[N][N], f[N][N];
vector<int> odd, even;
int s, t;
vector<int> res[N];
bool dd[N];
int a[N];
bool findpath() {
  queue<int> Q;
  Q.push(s);
  for (int i = 0; i <= n + 1; i++) pa[i] = -1;
  while (!Q.empty()) {
    int u = Q.front();
    Q.pop();
    for (int v = 0; v <= n + 1; v++)
      if (pa[v] == -1 && f[u][v] < c[u][v]) {
        pa[v] = u;
        if (v == t) return 1;
        Q.push(v);
      }
  }
  return 0;
}
void incflow() {
  int u = t;
  int minEdge = 3;
  while (u != s) {
    minEdge = min(minEdge, c[pa[u]][u] - f[pa[u]][u]);
    u = pa[u];
  }
  MaxFlow += minEdge;
  u = t;
  while (u != s) {
    f[pa[u]][u] += minEdge;
    f[u][pa[u]] -= minEdge;
    u = pa[u];
  }
}
void DFS(int u, int k) {
  dd[u] = 1;
  res[k].push_back(u);
  for (int i = 0; i < ed2[u].size(); i++) {
    int v = ed2[u][i];
    if (!dd[v]) {
      DFS(v, k);
      break;
    }
  }
}
const int NMAX = 100000 * 2;
bool NotPrime[NMAX + 3];
void Init() {
  NotPrime[1] = 1;
  for (int i = 2; i <= NMAX; i++)
    if (!NotPrime[i]) {
      for (int j = 2; j * i <= NMAX; j++) NotPrime[i * j] = 1;
    }
}
int main() {
  scanf("%d", &n);
  Init();
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] & 1)
      odd.push_back(i);
    else
      even.push_back(i);
  }
  if (odd.size() * 2 != n) return 0 * printf("Impossible");
  s = 0;
  t = n + 1;
  for (int i = 0; i < odd.size(); i++) c[s][i + 1] = 2;
  for (int i = 0; i < even.size(); i++) c[i + odd.size() + 1][t] = 2;
  for (int i = 0; i < odd.size(); i++)
    for (int j = 0; j < even.size(); j++)
      if (!NotPrime[a[odd[i]] + a[even[j]]]) {
        c[i + 1][j + odd.size() + 1] = 1;
      }
  while (findpath()) incflow();
  if (MaxFlow < n) return 0 * printf("Impossible");
  cnt = 0;
  for (int i = 0; i < odd.size(); i++)
    for (int j = 0; j < even.size(); j++)
      if (f[i + 1][j + odd.size() + 1] == 1)
        ed2[i + 1].push_back(j + odd.size() + 1),
            ed2[j + odd.size() + 1].push_back(i + 1);
  for (int i = 0; i < odd.size(); i++)
    if (!dd[i + 1]) {
      cnt++;
      DFS(i + 1, cnt);
    }
  printf("%d\n", cnt);
  for (int i = 1; i <= cnt; i++) {
    printf("%d ", res[i].size());
    for (int j = 0; j < res[i].size(); j++) {
      if (res[i][j] <= odd.size())
        printf("%d ", odd[res[i][j] - 1]);
      else
        printf("%d ", even[res[i][j] - odd.size() - 1]);
    }
    printf("\n");
  }
}
