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 INF = 0x3f3f3f3f;
struct Edge {
  int from, to, cap, flow;
  Edge(int u, int v, int c, int f) : from(u), to(v), cap(c), flow(f) {}
};
vector<Edge> edges;
vector<int> G[1010];
bool vis[1010];
int d[1010];
int cur[1010];
int n, m, s, t;
int num[1010];
vector<int> g[1010];
vector<int> ans[1010];
bool f[1010];
int block;
void init() {
  for (int i = 0; i < 1010; i++) G[i].clear();
  edges.clear();
}
void AddEdge(int from, int to, int cap) {
  edges.push_back(Edge(from, to, cap, 0));
  edges.push_back(Edge(to, from, 0, 0));
  int w = edges.size();
  G[from].push_back(w - 2);
  G[to].push_back(w - 1);
}
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) {
      edges[G[x][i]].flow += f;
      edges[G[x][i] ^ 1].flow -= f;
      flow += f;
      a -= f;
      if (a == 0) break;
    }
  }
  if (!flow) d[x] = -1;
  return flow;
}
int dinic(int s, int t) {
  int flow = 0;
  while (BFS()) {
    memset(cur, 0, sizeof(cur));
    flow += DFS(s, INF);
  }
  return flow;
}
bool prime(int x) {
  for (int i = 2; i * i <= x; i++)
    if (x % i == 0) return 0;
  return 1;
}
void Find(int now) {
  f[now] = 1;
  ans[block].push_back(now);
  for (int i = 0; i < g[now].size(); i++) {
    if (f[g[now][i]]) continue;
    Find(g[now][i]);
  }
}
int main() {
  while (~scanf("%d", &n)) {
    for (int i = 1; i <= n; i++) scanf("%d", &num[i]);
    init();
    s = 0;
    t = n + 1;
    int e1 = 0, e2 = 0;
    int flag = 0;
    for (int i = 1; i <= n; i++)
      if (num[i] % 2 == 0) {
        e1++;
        AddEdge(s, i, 2);
      } else {
        e2++;
        AddEdge(i, t, 2);
      }
    if (e1 != e2)
      flag = 1;
    else {
      for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
          if (num[i] % 2 == 0 && num[j] % 2 == 1 && prime(num[i] + num[j]))
            AddEdge(i, j, 1);
      int Flow = dinic(s, t);
      if (Flow != 2 * e1) flag = 1;
    }
    if (flag == 1)
      printf("Impossible\n");
    else {
      block = 0;
      memset(f, 0, sizeof f);
      for (int i = 0; i < 1010; i++) {
        g[i].clear();
        ans[i].clear();
      }
      for (int i = 0; i < edges.size(); i = i + 2) {
        if (edges[i].flow == 1) {
          int u = edges[i].from;
          int v = edges[i].to;
          g[u].push_back(v);
          g[v].push_back(u);
        }
      }
      for (int i = 1; i <= n; i++) {
        if (f[i]) continue;
        Find(i);
        block++;
      }
      printf("%d\n", block);
      for (int i = 0; i < block; i++) {
        printf("%d ", ans[i].size());
        for (int j = 0; j < ans[i].size(); j++) printf("%d ", ans[i][j]);
        printf("\n");
      }
    }
  }
  return 0;
}
