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;
const int MAXN = 2e2 + 5;
const int INF = 1e9;
int N;
int a[MAXN];
int edge[MAXN][MAXN];
vector<int> adj[MAXN], fin[MAXN];
vector<vector<int> > ans;
vector<int> comp;
bool bio[MAXN];
int cnt[MAXN];
int flow;
void load() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) scanf("%d", a + i);
}
bool prime(int x) {
  for (int i = 2; i * i <= x; i++)
    if (!(x % i)) return false;
  return true;
}
void add(int x, int y) {
  adj[x].push_back(y);
  adj[y].push_back(x);
  if (a[x] % 2) swap(x, y);
  edge[x][y] = 1;
}
int dfs(int x, int cap) {
  if (cap <= 0) return 0;
  bio[x] = true;
  if (x == N + 1) {
    flow += cap;
    return cap;
  }
  for (auto it : adj[x])
    if (!bio[it]) {
      int tmp = dfs(it, min(cap, edge[x][it]));
      if (tmp > 0) {
        edge[x][it] -= tmp;
        edge[it][x] += tmp;
        return tmp;
      }
    }
  return 0;
}
void go(int x) {
  bio[x] = true;
  comp.push_back(x);
  for (auto it : fin[x])
    if (!bio[it]) go(it);
}
bool check(const vector<int> &v) {
  if (v.size() < 3) return false;
  for (int i = 0; i < v.size(); i++) {
    if (!prime(a[v[i]] + a[v[(i + 1) % v.size()]])) return false;
    cnt[v[i]]++;
  }
  return true;
}
void solve() {
  for (int i = 1; i <= N; i++) {
    adj[0].push_back(i);
    adj[i].push_back(N + 1);
    if (a[i] % 2)
      edge[i][N + 1] = 2;
    else
      edge[0][i] = 2;
  }
  for (int i = 1; i <= N; i++)
    for (int j = i + 1; j <= N; j++)
      if (prime(a[i] + a[j])) add(i, j);
  do memset(bio, false, sizeof bio);
  while (dfs(0, INF));
  if (flow != N) {
    puts("Impossible");
    return;
  }
  for (int i = 1; i <= N; i++)
    if (!(a[i] % 2))
      for (auto it : adj[i])
        if (it >= 1 && it <= N && !edge[i][it]) {
          fin[i].push_back(it);
          fin[it].push_back(i);
        }
  memset(bio, false, sizeof bio);
  for (int i = 1; i <= N; i++)
    if (!bio[i]) {
      comp.clear();
      go(i);
      ans.push_back(comp);
    }
  printf("%d\n", ans.size());
  for (auto cyc : ans) {
    assert(check(cyc));
    printf("%d", cyc.size());
    for (auto it : cyc) printf(" %d", it);
    puts("");
  }
  for (int i = 1; i <= N; i++) assert(cnt[i] == 1);
}
int main() {
  load();
  solve();
  return 0;
}