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 MAX_V = 220;
const int MAX_E = MAX_V * MAX_V;
int to[MAX_E], cap[MAX_E], ant[MAX_E], z;
int adj[MAX_V], copy_adj[MAX_V], fila[MAX_V], level[MAX_V];
int bfs(int source, int sink) {
  memset(level, -1, sizeof level);
  level[source] = 0;
  int front = 0, size = 0;
  fila[size++] = source;
  while (front < size) {
    int v = fila[front++];
    for (int i = adj[v]; ~i; i = ant[i]) {
      if (cap[i] && level[to[i]] == -1) {
        level[to[i]] = level[v] + 1;
        fila[size++] = to[i];
      }
    }
  }
  return level[sink] != -1;
}
int dfs(int v, int sink, int flow) {
  if (v == sink) return flow;
  int f;
  for (int& i = copy_adj[v]; ~i; i = ant[i]) {
    if (cap[i] && level[to[i]] == level[v] + 1 &&
        (f = dfs(to[i], sink, min(flow, cap[i])))) {
      cap[i] -= f, cap[i ^ 1] += f;
      return f;
    }
  }
  return 0;
}
int maxflow(int source, int sink) {
  int ret = 0, flow;
  while (bfs(source, sink)) {
    memcpy(copy_adj, adj, sizeof adj);
    while ((flow = dfs(source, sink, 1 << 30))) ret += flow;
  }
  return ret;
}
bool prime[20010];
int a[222];
vector<vector<int>> ans;
bool mark[222];
int source, sink;
void cycle(int u) {
  mark[u] = true;
  ans.back().push_back(u + 1);
  for (int x = adj[u]; ~x; x = ant[x]) {
    int v = to[x];
    if (v == source || v == sink) continue;
    if (a[v] % 2 == 0 && cap[x] != 0) continue;
    if (a[v] % 2 == 1 && cap[x ^ 1] != 0) continue;
    if (!mark[v]) {
      cycle(v);
    }
  }
}
int main() {
  memset(adj, -1, sizeof(adj));
  z = 0;
  memset(prime, 1, sizeof prime);
  prime[0] = prime[1] = 0;
  for (int i = 2; i < 20010; ++i)
    if (prime[i]) {
      for (int j = i + i; j < 20010; j += i) {
        prime[j] = 0;
      }
    }
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
  source = n, sink = n + 1;
  int odd = 0, even = 0;
  for (int i = 0; i < n; ++i) {
    if (a[i] & 1) {
      to[z] = i, cap[z] = 2, ant[z] = adj[source], adj[source] = z++;
      to[z] = source, cap[z] = 0, ant[z] = adj[i], adj[i] = z++;
      ++odd;
    } else {
      to[z] = sink, cap[z] = 2, ant[z] = adj[i], adj[i] = z++;
      to[z] = i, cap[z] = 0, ant[z] = adj[sink], adj[sink] = z++;
      ++even;
    }
  }
  for (int i = 0; i < n; ++i)
    if (a[i] & 1) {
      for (int j = 0; j < n; ++j)
        if (a[j] % 2 == 0) {
          if (prime[a[i] + a[j]]) {
            to[z] = j, cap[z] = 1, ant[z] = adj[i], adj[i] = z++;
            to[z] = i, cap[z] = 0, ant[z] = adj[j], adj[j] = z++;
          }
        }
    }
  int mf = maxflow(source, sink);
  if (odd != even || mf != 2 * odd) {
    printf("Impossible\n");
    return 0;
  } else {
    for (int x = adj[source]; ~x; x = ant[x]) {
      assert(cap[x] == 0);
      int v = to[x];
      if (!mark[v]) {
        ans.push_back(vector<int>());
        cycle(v);
        if (ans.back().size() < 3) {
          printf("Impossible\n");
          return 0;
        }
      }
    }
  }
  printf("%d\n", int(ans.size()));
  for (const auto& v : ans) {
    printf("%d ", int(v.size()));
    for (int x : v) {
      printf("%d ", x);
    }
    printf("\n");
  }
  return 0;
}
