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 to, cap, flow, rev;
  edge(int _to, int _cap, int _flow, int _rev) {
    to = _to, cap = _cap, flow = _flow, rev = _rev;
  }
};
struct Dinic {
  vector<edge> adj[305];
  int level[305], start[305];
  int sink, src;
  Dinic(int s, int t) {
    src = s;
    sink = t;
    memset(start, 0, sizeof(start));
  }
  void addEdge(int from, int to, int cap1, int cap2) {
    edge e1 = edge(to, cap1, 0, adj[to].size());
    edge e2 = edge(from, cap2, 0, adj[from].size());
    adj[from].push_back(e1);
    adj[to].push_back(e2);
  }
  bool bfs() {
    memset(level, -1, sizeof(level));
    level[src] = 0;
    queue<int> q;
    q.push(src);
    while (!q.empty()) {
      int cur = q.front();
      q.pop();
      for (edge &e : adj[cur]) {
        if (level[e.to] == -1 && e.flow < e.cap) {
          level[e.to] = level[cur] + 1;
          q.push(e.to);
        }
      }
    }
    return level[sink] != -1;
  }
  int sendFlow(int u, int flow) {
    if (u == sink) return flow;
    for (; start[u] < ((int)adj[u].size()); start[u]++) {
      edge &e = adj[u][start[u]];
      if (level[e.to] == level[u] + 1 && e.flow < e.cap) {
        int curflow = min(flow, e.cap - e.flow);
        curflow = sendFlow(e.to, curflow);
        if (curflow > 0) {
          e.flow += curflow;
          adj[e.to][e.rev].flow -= curflow;
          return curflow;
        }
      }
    }
    return 0;
  }
  int maxFlow() {
    if (src == sink) return 0;
    int totalflow = 0;
    while (bfs()) {
      memset(start, 0, sizeof(start));
      int flow;
      while ((flow = sendFlow(src, INT_MAX))) totalflow += flow;
    }
    return totalflow;
  }
};
vector<bool> isPrime;
int n, arr[205];
vector<int> adj[305];
bool vis[305];
vector<vector<int> > ans;
void sieve(int limit) {
  isPrime.assign(limit, 1);
  isPrime[0] = isPrime[1] = 0;
  for (int i = 2; i * i <= limit; i++)
    if (isPrime[i])
      for (int j = i + i; j <= limit; j += i) isPrime[j] = 0;
}
void dfs(int node) {
  ans.back().push_back(node);
  vis[node] = 1;
  for (int child : adj[node]) {
    if (vis[child]) continue;
    dfs(child);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", arr + i);
  int source = 0, sink = n + 1;
  Dinic dinic(source, sink);
  for (int i = 1; i <= n; i++) {
    if (arr[i] % 2 == 0)
      dinic.addEdge(source, i, 2, 0);
    else
      dinic.addEdge(i, sink, 2, 0);
  }
  sieve(1e5);
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      if (!isPrime[arr[i] + arr[j]]) continue;
      if (arr[i] % 2 == 0)
        dinic.addEdge(i, j, 1, 0);
      else
        dinic.addEdge(j, i, 1, 0);
    }
  }
  int r = dinic.maxFlow();
  if (r != n) {
    puts("Impossible");
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    if (arr[i] % 2) continue;
    for (edge &e : dinic.adj[i]) {
      if (e.to >= 1 && e.to <= n && e.flow == 1) {
        adj[i].push_back(e.to);
        adj[e.to].push_back(i);
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    if (vis[i]) continue;
    ans.push_back(vector<int>());
    dfs(i);
  }
  printf("%d\n", ((int)ans.size()));
  for (int i = 0; i < ((int)ans.size()); i++) {
    printf("%d", ((int)ans[i].size()));
    for (int j = 0; j < ((int)ans[i].size()); j++) printf(" %d", ans[i][j]);
    puts("");
  }
  return 0;
}