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>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
struct Edge {
  int from, to, cap, nex;
} edge[100100 * 2];
int head[500], tot;
void add(int u, int v, int cap, int rw = 0) {
  Edge E = {u, v, cap, head[u]};
  edge[tot] = E;
  head[u] = tot++;
  Edge E2 = {v, u, rw, head[v]};
  edge[tot] = E2;
  head[v] = tot++;
}
int dis[500], cur[500];
bool vis[500];
bool BFS(int Start, int End) {
  memset(vis, 0, sizeof(vis));
  memset(dis, -1, sizeof(dis));
  queue<int> Q;
  Q.push(Start);
  dis[Start] = 0;
  vis[Start] = 1;
  while (!Q.empty()) {
    int u = Q.front();
    Q.pop();
    for (int i = head[u]; i != -1; i = edge[i].nex) {
      Edge E = edge[i];
      if (!vis[E.to] && E.cap > 0) {
        vis[E.to] = true;
        dis[E.to] = dis[u] + 1;
        if (E.to == End) return true;
        Q.push(E.to);
      }
    }
  }
  return false;
}
int DFS(int x, int a, int End) {
  if (x == End || a == 0) return a;
  int flow = 0, f;
  for (int& i = cur[x]; i != -1; i = edge[i].nex) {
    Edge& E = edge[i];
    if (dis[x] + 1 == dis[E.to] && (f = DFS(E.to, min(a, E.cap), End)) > 0) {
      E.cap -= f;
      edge[i ^ 1].cap += f;
      flow += f;
      a -= f;
      if (a == 0) break;
    }
  }
  return flow;
}
int Dinic(int Start, int End) {
  int flow = 0;
  while (BFS(Start, End)) {
    memcpy(cur, head, sizeof(head));
    flow += DFS(Start, 1073741824, End);
  }
  return flow;
}
void init() {
  memset(head, -1, sizeof head);
  tot = 0;
}
int prime[10000], primenum;
set<int> p;
void PRIME(int Max_Prime) {
  p.clear();
  primenum = 0;
  prime[primenum++] = 2;
  for (int i = 3; i <= Max_Prime; i += 2)
    for (int j = 0; j < primenum; j++)
      if (i % prime[j] == 0)
        break;
      else if (prime[j] > sqrt((double)i) || j == primenum - 1) {
        prime[primenum++] = i;
        break;
      }
  for (int i = 0; i < primenum; i++) p.insert(prime[i]);
}
int a[500], n;
vector<int> G[500];
int top;
bool Vis[500];
void dfs(int u, vector<int>& x) {
  Vis[u] = 1;
  x.push_back(u);
  for (int i = head[u]; ~i; i = edge[i].nex) {
    int v = edge[i].to;
    if (Vis[v]) continue;
    if (a[u] & 1) {
      if (edge[i].cap != 1) {
        edge[i].cap++;
        dfs(v, x);
      }
    } else {
      if (edge[i].cap) {
        edge[i].cap--;
        dfs(v, x);
      }
    }
  }
}
int main() {
  PRIME(20000);
  while (cin >> n) {
    for (int i = 0; i < n; i++) cin >> a[i];
    if (n & 1) {
      puts("Impossible");
      continue;
    }
    init();
    int from = n, to = n + 1;
    Vis[from] = Vis[to] = 1;
    for (int i = 0; i < n; i++) {
      G[i].clear();
      Vis[i] = 0;
      if (a[i] & 1)
        add(from, i, 2);
      else
        add(i, to, 2);
      for (int j = 0; j < n; j++)
        if (p.count(a[i] + a[j])) {
          if (a[i] & 1) add(i, j, 1);
        }
    }
    int all = Dinic(from, to);
    if (all == n) {
      top = 0;
      for (int i = 0; i < n; i++)
        if (Vis[i] == 0) {
          dfs(i, G[top++]);
        }
      printf("%d\n", top);
      for (int i = 0; i < top; i++) {
        printf("%d ", G[i].size());
        for (int j = 0; j < G[i].size(); j++) printf("%d ", G[i][j] + 1);
        puts("");
      }
    } else
      puts("Impossible");
  }
  return 0;
}
