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;
const int MAX = 1e5 + 40;
const int MAXE = 3e5 + 400;
const int MAXN = 1000 + 50;
const int INF = 1e9 + 5;
int n, ec, cnt;
int from[2 * MAXE], to[2 * MAXE], cap[2 * MAXE], prv[2 * MAXE];
int head[MAXN], age[MAXN];
bool mark[MAXN];
vector<int> od, ev, ans[MAXN], sec[MAXN], adj[MAXN];
bool pr[MAX];
int fs[2 * MAXE];
void init() {
  pr[0] = pr[1] = 1;
  for (int i = 2; i < MAX; i++)
    for (int j = i + i; j < MAX; j += i) pr[j] = 1;
  memset(head, -1, sizeof(mark));
}
void add_edge(int u, int v, int uv, int vu = 0) {
  fs[ec] = uv;
  from[ec] = u, to[ec] = v, cap[ec] = uv, prv[ec] = head[u];
  head[u] = ec++;
  fs[ec] = vu;
  from[ec] = v, to[ec] = u, cap[ec] = vu, prv[ec] = head[v], head[v] = ec++;
}
int dfs(int v, int sink, int mini = INF) {
  if (v == sink) return mini;
  if (mark[v]) return 0;
  mark[v] = 1;
  for (int e = head[v]; e != -1; e = prv[e])
    if (cap[e]) {
      int x = dfs(to[e], sink, min(mini, cap[e]));
      if (x == 0) continue;
      cap[e] -= x;
      cap[e ^ 1] += x;
      return x;
    }
  return 0;
}
int max_flow(int root, int sink) {
  memset(mark, 0, sizeof(mark));
  int ret = 0, x;
  while ((x = dfs(root, sink))) {
    ret += x;
    memset(mark, 0, sizeof(mark));
  }
  return ret;
}
void get(int v) {
  if (mark[v]) return;
  mark[v] = 1;
  ans[cnt].push_back(v);
  for (int u : sec[v]) get(u);
}
int main() {
  init();
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> age[i];
    if (age[i] % 2 == 0)
      ev.push_back(i);
    else
      od.push_back(i);
  }
  for (int i = 0; i < ev.size(); i++)
    for (int j = 0; j < od.size(); j++)
      if (!pr[age[ev[i]] + age[od[j]]]) add_edge(ev[i], od[j], 1);
  int temp = ec;
  for (int u : ev) add_edge(n, u, 2);
  for (int u : od) add_edge(u, n + 1, 2);
  if (od.size() != ev.size() || max_flow(n, n + 1) != 2 * (int)ev.size())
    cout << "Impossible\n";
  else {
    for (int i = 0; i < temp; i += 2)
      if (fs[i] != cap[i]) {
        sec[from[i]].push_back(to[i]);
        sec[to[i]].push_back(from[i]);
      }
    memset(mark, 0, sizeof(mark));
    for (int i = 0; i < n; i++)
      if (!mark[i]) get(i), cnt++;
    cout << cnt << endl;
    for (int i = 0; i < cnt; i++) {
      cout << ans[i].size() << " ";
      for (int u : ans[i]) cout << u + 1 << " ";
      cout << endl;
    }
  }
  return 0;
}