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;
struct edge {
  int v, f, nxt;
} e[50001];
int n, nume = 1, top, src, sink, a[202], h[202], g[202], dist[202], que[202];
bool vis[202];
vector<int> ans[201];
void add_edge(int u, int v, int f) {
  e[++nume].v = v, e[nume].f = f, e[nume].nxt = h[u], h[u] = nume;
  e[++nume].v = u, e[nume].f = 0, e[nume].nxt = h[v], h[v] = nume;
}
void bfs() {
  memset(dist, 0, sizeof(dist));
  que[1] = src, dist[src] = 1;
  int s = 0, t = 1;
  while (s < t) {
    int u = que[++s];
    for (int i = h[u]; i; i = e[i].nxt) {
      if (e[i].f && !dist[e[i].v]) {
        que[++t] = e[i].v, dist[e[i].v] = dist[u] + 1;
      }
    }
  }
}
int dfs(int u, int delta) {
  if (u == sink) return delta;
  int ret = 0;
  for (int i = g[u]; i; i = e[i].nxt) {
    if (e[i].f && dist[e[i].v] == dist[u] + 1) {
      int dd = dfs(e[i].v, min(e[i].f, delta));
      e[i].f -= dd, e[i ^ 1].f += dd;
      if (e[i].f) g[u] = i;
      delta -= dd, ret += dd;
    }
  }
  return ret;
}
int max_flow() {
  int ret = 0;
  while (1) {
    bfs();
    if (!dist[sink]) return ret;
    for (int i = 0; i <= n + 1; ++i) g[i] = h[i];
    ret += dfs(src, 1e9);
  }
}
bool is_prime(int t) {
  int k = (int)sqrt(t);
  for (int i = 2; i <= k; ++i) {
    if (!(t % i)) return false;
  }
  return true;
}
void find(int u, int t) {
  vis[u] = true, ans[t].push_back(u);
  for (int i = h[u]; i; i = e[i].nxt) {
    if (e[i].v && e[i].v != n + 1 && e[i].f == !(a[u] & 1) && !vis[e[i].v]) {
      find(e[i].v, t);
    }
  }
}
int main() {
  cin >> n;
  src = 0, sink = n + 1;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    if (a[i] & 1)
      add_edge(src, i, 2);
    else
      add_edge(i, sink, 2);
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      if (a[i] & 1 && !(a[j] & 1) && is_prime(a[i] + a[j])) add_edge(i, j, 1);
    }
  }
  max_flow();
  for (int i = h[0]; i; i = e[i].nxt) {
    if (e[i].f) {
      cout << "Impossible" << endl;
      return 0;
    }
  }
  for (int i = h[n + 1]; i; i = e[i].nxt) {
    if (e[i].f < 2) {
      cout << "Impossible" << endl;
      return 0;
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (!vis[i]) top++, find(i, top);
  }
  cout << top << endl;
  for (int i = 1; i <= top; ++i) {
    cout << ans[i].size() << ' ';
    for (vector<int>::iterator iter = ans[i].begin(); iter != ans[i].end();
         ++iter) {
      cout << *iter << ' ';
    }
    cout << endl;
  }
  return 0;
}