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;
vector<long long> v[250];
long long cap[250][250];
long long res[250][250];
long long par[250];
long long act[250];
long long a[250];
long long find(long long sour, long long sink) {
  long long curr;
  long long ans = 0;
  long long from;
  while (1) {
    memset(par, -1, sizeof par);
    queue<int> q;
    q.push(sour);
    par[sour] = 0;
    act[sour] = 1e18;
    while (!q.empty()) {
      curr = q.front();
      q.pop();
      for (int i = 0; i < v[curr].size(); i++) {
        if (par[v[curr][i]] == -1 &&
            cap[curr][v[curr][i]] - res[curr][v[curr][i]] > 0) {
          act[v[curr][i]] =
              min(act[curr], cap[curr][v[curr][i]] - res[curr][v[curr][i]]);
          par[v[curr][i]] = curr;
          q.push(v[curr][i]);
          if (v[curr][i] == sink) {
            ans += (act[sink]);
            break;
          }
        }
      }
    }
    if (par[sink] == -1) {
      break;
    } else {
      curr = sink;
      while (par[curr] != 0) {
        from = par[curr];
        res[from][curr] += act[sink];
        res[curr][from] -= act[sink];
        curr = from;
      }
    }
  }
  return ans;
}
bool prime[20010];
void sieve() {
  memset(prime, true, sizeof prime);
  prime[1] = false;
  for (int i = 4; i <= 20000; i += 2) prime[i] = false;
  for (int i = 3; i * i <= 20000; i += 2) {
    if (prime[i]) {
      for (int j = 2 * i; j <= 20000; j += i) {
        prime[j] = false;
      }
    }
  }
}
vector<long long> table[200];
bool visit[250];
int main() {
  sieve();
  long long n, sour, sink;
  cin >> n;
  sour = n + 1;
  sink = n + 2;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  long long start, end;
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      if (prime[a[i] + a[j]]) {
        if (a[i] & 1) {
          start = j;
          end = i;
        } else {
          start = i;
          end = j;
        }
        v[start].push_back(end);
        cap[start][end] = 1;
        v[end].push_back(start);
        cap[end][start] = 0;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    if (a[i] & 1) {
      v[i].push_back(sink);
      cap[i][sink] = 2;
    } else {
      v[sour].push_back(i);
      cap[sour][i] = 2;
    }
  }
  long long ta = 0;
  long long curr;
  memset(visit, false, sizeof visit);
  if (find(sour, sink) == n) {
    for (int i = 1; i <= n; i++) {
      if (!visit[i]) {
        curr = i;
        while (!visit[curr]) {
          table[ta].push_back(curr);
          visit[curr] = true;
          for (int j = 1; j <= n; j++) {
            if (!visit[j] && abs(res[curr][j]) == 1) {
              curr = j;
              break;
            }
          }
        }
        ta++;
      }
    }
    cout << ta << endl;
    for (int i = 0; i < ta; i++) {
      cout << table[i].size() << " ";
      for (int j = 0; j < table[i].size(); j++) {
        cout << table[i][j] << " ";
      }
      cout << endl;
    }
  } else {
    cout << "Impossible" << endl;
  }
  return 0;
}
