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 MAXN = 300;
const int MAXP = 100000;
const int INF = 100000;
struct edge {
  int a, b, cap, flow;
};
vector<edge> e;
vector<int> g[MAXN];
int dist[MAXN], ptr[MAXN], a[MAXN];
queue<int> q;
int s, t;
bool used[MAXN];
vector<int> ans;
void Add(int u, int v, int cap) {
  edge e1 = {u, v, cap, 0};
  edge e2 = {v, u, 0, 0};
  g[u].push_back(e.size());
  e.push_back(e1);
  g[v].push_back(e.size());
  e.push_back(e2);
}
bool bfs() {
  memset(dist, -1, sizeof(dist));
  int i, v, to;
  dist[s] = 0;
  q.push(s);
  while ((!q.empty()) && (dist[t] == -1)) {
    v = q.front();
    q.pop();
    for (i = 0; i < g[v].size(); i++) {
      to = e[g[v][i]].b;
      if (dist[to] != -1) continue;
      if (e[g[v][i]].cap > e[g[v][i]].flow) {
        dist[to] = dist[v] + 1;
        q.push(to);
      }
    }
  }
  while (!q.empty()) q.pop();
  return (dist[t] != -1);
}
int dfs(int v, int flow) {
  int i, id, to, pushed, pt;
  if (!flow) return 0;
  if (v == t) return flow;
  pt = ptr[v];
  for (i = pt; i < g[v].size(); i++) {
    ptr[v]++;
    id = g[v][i];
    to = e[id].b;
    if ((dist[v] + 1) != dist[to]) continue;
    pushed = dfs(to, min(flow, e[id].cap - e[id].flow));
    if (pushed) {
      e[id].flow += pushed;
      e[id ^ 1].flow -= pushed;
      return pushed;
    }
  }
  return 0;
}
int dinic() {
  int flow = 0, pushed;
  for (;;) {
    if (!bfs()) break;
    memset(ptr, 0, sizeof(ptr));
    while (pushed = dfs(s, INF)) flow += pushed;
  }
  return flow;
}
void GetAns(int v) {
  int i, to;
  ans.push_back(v - 1);
  used[v] = true;
  for (i = 0; i < g[v].size(); i++) {
    to = e[g[v][i]].b;
    if ((to == s) || (to == t)) continue;
    if (used[to]) continue;
    if (e[g[v][i]].flow != 0) GetAns(to);
  }
}
int main() {
  int n, i, j, even, cnt;
  bool pr[MAXP];
  memset(pr, true, sizeof(pr));
  for (i = 2; i < MAXP; i++) {
    if (!pr[i]) continue;
    for (j = 2; j * i < MAXP; j++) pr[i * j] = false;
  }
  cin >> n;
  even = 0;
  s = 1;
  t = n + 2;
  for (i = 2; i <= n + 1; i++) {
    cin >> a[i];
    if (a[i] & 1) {
      Add(i, t, 2);
      even++;
    } else
      Add(s, i, 2);
  }
  if (((n & 1) == 1) || (even != n / 2)) {
    cout << "Impossible" << endl;
    return 0;
  }
  for (i = 2; i <= n + 1; i++) {
    for (j = i + 1; j <= n + 1; j++) {
      if (!pr[a[i] + a[j]]) continue;
      if (a[i] & 1)
        Add(j, i, 1);
      else
        Add(i, j, 1);
    }
  }
  if (2 * even != dinic()) {
    cout << "Impossible" << endl;
    return 0;
  }
  memset(used, false, sizeof(used));
  cnt = 0;
  for (i = 2; i <= n + 1; i++) {
    if (used[i]) continue;
    cnt++;
    ans.clear();
    GetAns(i);
  }
  cout << cnt << endl;
  memset(used, false, sizeof(used));
  for (i = 2; i <= n + 1; i++) {
    if (used[i]) continue;
    ans.clear();
    GetAns(i);
    cout << ans.size();
    for (j = 0; j < ans.size(); j++) cout << " " << ans[j];
    cout << endl;
  }
  return 0;
}
