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;
long long to[51000], ref[300], cap[510000], ant[5100000], adj[300], z;
int n, m, brs;
void insert(int a, int b, long long c = 1) {
  ant[z] = adj[a];
  cap[z] = c;
  to[z] = b;
  adj[a] = z++;
  ant[z] = adj[b];
  cap[z] = 0;
  to[z] = a;
  adj[b] = z++;
}
long long level[300], fila[900000], copy_adj[300];
int bfs(int source, int sink) {
  memset(level, -1, sizeof level);
  level[source] = 0;
  int pos = 0, tam = 0;
  fila[tam++] = source;
  while (pos < tam) {
    int now = fila[pos++];
    for (int i = adj[now]; ~i; i = ant[i]) {
      if (cap[i] && level[to[i]] == -1) {
        level[to[i]] = level[now] + 1;
        fila[tam++] = to[i];
      }
    }
  }
  return ~level[sink];
}
long long dfs(int no, int sink, long long flow) {
  if (no == sink) return flow;
  for (long long &i = copy_adj[no]; ~i; i = ant[i]) {
    if (cap[i] && level[no] + 1 == level[to[i]]) {
      long long nflow = dfs(to[i], sink, min(flow, cap[i]));
      if (nflow) {
        cap[i] -= nflow;
        cap[i ^ 1] += nflow;
        return nflow;
      }
    }
  }
  return 0;
}
long long dinic(int source, int sink) {
  long long mf = 0, add;
  while (bfs(source, sink)) {
    memcpy(copy_adj, adj, sizeof adj);
    while (add = dfs(source, sink, 10000)) {
      mf += add;
    }
  }
  return mf;
}
bitset<20005> mkd;
void sieve() {
  mkd.reset();
  for (int i = 2; i < 1000; i++)
    if (!mkd[i])
      for (int j = i * i; j < 20001; j += i) mkd[j] = 1;
}
int lido[300];
int mkc[300];
vector<int> vet;
vector<vector<int> > ans;
void dfsP(int n) {
  mkd[n] = 1;
  vet.push_back(n - 1);
  for (int e = adj[n]; ~e; e = ant[e]) {
    if (!mkd[to[e]] && cap[e | 1] && to[e] > 1) dfsP(to[e]);
  }
}
void matchmake() {
  int k = 2;
  sieve();
  memset(adj, -1, sizeof adj);
  z = 0;
  for (int i = 0; i < n; i++) {
    if (lido[i] & 1)
      insert(i + 2, 1, 2);
    else
      insert(0, i + 2, 2);
  }
  for (int i = 0; i < n; i++)
    if (lido[i] % 2 == 0) {
      for (int j = 0; j < n; j++)
        if (i != j) {
          if (!mkd[lido[i] + lido[j]]) insert(2 + i, j + 2);
        }
    }
  if (dinic(0, 1) < n) {
    cout << "Impossible\n";
    return;
  }
  mkd.reset();
  for (int i = 0; i < n; i++) {
    vet.clear();
    if (!mkd[i + 2]) {
      dfsP(i + 2);
      ans.push_back(vet);
    }
  }
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i].size();
    for (int j = 0; j < ans[i].size(); j++) cout << " " << ans[i][j];
    cout << endl;
  }
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> lido[i];
  matchmake();
}
