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;
int prime[20101];
const int maxn = 2001;
int a[maxn];
int Vis[maxn];
struct node {
  int x;
  int id;
} odd[maxn], even[maxn];
vector<int> ans[maxn];
struct Edge {
  int from, to, cap, flow;
  Edge(int u, int v, int c, int f) : from(u), to(v), cap(c), flow(f) {}
};
struct Dinic {
  int n, m, s, t;
  vector<Edge> edges;
  vector<int> G[maxn];
  bool vis[maxn];
  int d[maxn];
  int cur[maxn];
  void AddEdge(int from, int to, int cap) {
    edges.push_back(Edge(from, to, cap, 0));
    edges.push_back(Edge(to, from, 0, 0));
    m = edges.size();
    G[from].push_back(m - 2);
    G[to].push_back(m - 1);
  }
  void init(int n) {
    this->n = n;
    for (int i = 0; i <= n; i++) G[i].clear();
    edges.clear();
  }
  bool BFS() {
    memset(vis, 0, sizeof(vis));
    queue<int> Q;
    Q.push(s);
    d[s] = 0;
    vis[s] = 1;
    while (!Q.empty()) {
      int x = Q.front();
      Q.pop();
      for (int i = 0; i < G[x].size(); i++) {
        Edge& e = edges[G[x][i]];
        if (!vis[e.to] && e.cap > e.flow) {
          vis[e.to] = 1;
          d[e.to] = d[x] + 1;
          Q.push(e.to);
        }
      }
    }
    return vis[t];
  }
  int DFS(int x, int a) {
    if (x == t || a == 0) return a;
    int flow = 0, f;
    for (int& i = cur[x]; i < G[x].size(); i++) {
      Edge& e = edges[G[x][i]];
      if (d[x] + 1 == d[e.to] && (f = DFS(e.to, min(a, e.cap - e.flow))) > 0) {
        e.flow += f;
        edges[G[x][i] ^ 1].flow -= f;
        flow += f;
        a -= f;
        if (a == 0) break;
      }
    }
    return flow;
  }
  int Maxflow(int s, int t) {
    this->s = s;
    this->t = t;
    int flow = 0;
    while (BFS()) {
      memset(cur, 0, sizeof(cur));
      flow += DFS(s, 0x3f3f3f3f);
    }
    return flow;
  }
  void dfs1(int u, int top) {
    Vis[u] = 1;
    ans[top].push_back(u);
    for (int i = 0; i < G[u].size(); i++) {
      Edge& e = edges[G[u][i]];
      if (Vis[e.to] || e.to == 0 || e.to == n + 1) continue;
      if (a[u] & 1) {
        if (e.cap - e.flow == 0) {
          dfs1(e.to, top);
        }
      } else {
        if (e.cap - e.flow != 0) {
          dfs1(e.to, top);
        }
      }
    }
  }
};
Dinic solve;
int main() {
  int n;
  memset(prime, 0, sizeof(prime));
  for (int i = 2; i <= 20001; i++) {
    if (!prime[i])
      for (int j = i * i; j <= 20001; j += i) {
        prime[j] = 1;
      }
  }
  while (scanf("%d", &n) != EOF) {
    solve.init(n);
    for (int i = 0; i < n; i++) ans[i].clear();
    int cnt1 = 0, cnt2 = 0;
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      if (a[i] % 2 == 0) {
        even[cnt1].x = a[i];
        even[cnt1++].id = i;
        solve.AddEdge(i, n + 1, 2);
      } else {
        odd[cnt2].x = a[i];
        odd[cnt2++].id = i;
        solve.AddEdge(0, i, 2);
      }
    }
    if (n % 2 == 1) {
      printf("Impossible\n");
      continue;
    }
    for (int i = 0; i < cnt2; i++)
      for (int j = 0; j < cnt1; j++) {
        if (prime[odd[i].x + even[j].x] == 0) {
          solve.AddEdge(odd[i].id, even[j].id, 1);
        }
      }
    int flow = solve.Maxflow(0, n + 1);
    if (flow == n) {
      memset(Vis, 0, sizeof(Vis));
      int top = 0;
      for (int i = 1; i <= n; i++) {
        if (Vis[i] == 0) {
          solve.dfs1(i, top++);
        }
      }
      printf("%d\n", top);
      for (int i = 0; i < top; i++) {
        printf("%d", ans[i].size());
        for (int j = 0; j < ans[i].size(); j++) {
          printf(" %d", ans[i][j]);
        }
        printf("\n");
      }
    } else
      printf("Impossible\n");
  }
  return 0;
}