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>
namespace G {
int n, m, S, T;
struct Edge {
  int from, to, cap, nxt;
} e[100000 * 2 + 10];
int head[10000 + 10], cur[10000 + 10], cnt;
inline void Init() { memset(head, -1, sizeof(head)), cnt = -1; }
inline void AddEdge(int u, int v, int w) {
  e[++cnt] = (Edge){u, v, w, head[u]}, head[u] = cnt;
  e[++cnt] = (Edge){v, u, 0, head[v]}, head[v] = cnt;
}
int dep[10000 + 10];
bool Dinic_BFS() {
  std::queue<int> q;
  q.push(S);
  memset(dep, 0, sizeof(dep));
  dep[S] = 1;
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    for (int i = head[x]; ~i; i = e[i].nxt)
      if (e[i].cap > 0 && dep[e[i].to] == 0)
        dep[e[i].to] = dep[x] + 1, q.push(e[i].to);
  }
  return dep[T];
}
int Dinic_DFS(int u, int now) {
  if (u == T) return now;
  int ret = 0;
  for (int &i = cur[u]; ~i; i = e[i].nxt)
    if (dep[e[i].to] == dep[u] + 1 && e[i].cap) {
      int di = Dinic_DFS(e[i].to,
                         ((e[i].cap) < (now - ret) ? (e[i].cap) : (now - ret)));
      if (di) {
        e[i].cap -= di, e[i ^ 1].cap += di, ret += di;
        if (ret == now) break;
      }
    }
  return ret;
}
inline int Dinic() {
  int ret = 0;
  while (Dinic_BFS()) {
    memcpy(cur, head, sizeof(head));
    ret += Dinic_DFS(S, (1 << 30));
  }
  return ret;
}
bool vis[10000 + 10];
std::vector<int> cir[10000 + 10];
void DFS(int u, int t) {
  vis[u] = true, cir[t].push_back(u);
  for (int i = head[u]; ~i; i = e[i].nxt)
    if (!vis[e[i].to] && e[i | 1].cap == 1) DFS(e[i].to, t);
}
}  // namespace G
bool Prime(int t) {
  for (int i = 2; i * i <= t; i++)
    if (t % i == 0) return false;
  return true;
}
const int S = 10000 - 2, T = 10000 - 1;
int a[10000 + 10];
int main() {
  G::Init();
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] & 1)
      G::AddEdge(S, i, 2);
    else
      G::AddEdge(i, T, 2);
  }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (a[i] & 1 && !(a[j] & 1) && Prime(a[i] + a[j])) G::AddEdge(i, j, 1);
  G::S = S, G::T = T, G::n = n;
  if (G::Dinic() != n) {
    puts("Impossible");
    return 0;
  }
  int ans = 0;
  for (int i = 1; i <= n; i++)
    if (!G::vis[i]) G::DFS(i, ++ans);
  printf("%d\n", ans);
  for (int i = 1; i <= ans; i++) {
    printf("%d ", G::cir[i].size());
    for (auto t : G::cir[i]) printf("%d ", t);
    puts("");
  }
  return 0;
}