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 Nmax = 210;
int n, res, st, fi, cnt;
int t[Nmax];
int f[Nmax][Nmax], c[Nmax][Nmax];
bool k[Nmax], p[20100];
vector<int> v[Nmax], b[Nmax];
void eratos() {
  memset(p, true, sizeof(p));
  p[1] = false;
  for (long long i = 2; i <= 20000; ++i)
    if (p[i])
      for (long long j = i * i; j <= 20000; j += i) p[j] = false;
}
void dfs(int x) {
  k[x] = false;
  v[res].push_back(x);
  for (auto y : b[x])
    if (k[y]) dfs(y);
}
bool findpath() {
  queue<int> q;
  memset(k, true, sizeof(k));
  q.push(st);
  k[st] = false;
  while (q.size()) {
    int x = q.front();
    q.pop();
    for (int i = 1; i <= n + 2; ++i)
      if (k[i] && f[x][i] < c[x][i]) {
        k[i] = false;
        q.push(i);
        t[i] = x;
        if (i == fi) return true;
      }
  }
  return false;
}
void incflow() {
  cnt++;
  int x = fi;
  while (x != st) {
    f[t[x]][x]++;
    f[x][t[x]]--;
    x = t[x];
  }
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> t[i];
  eratos();
  for (int i = 1; i <= n; ++i)
    if (t[i] % 2 == 0) {
      for (int j = 1; j <= n; ++j)
        if (p[t[i] + t[j]]) c[i][j] = 1;
    }
  st = n + 1;
  fi = n + 2;
  for (int i = 1; i <= n; ++i)
    if (t[i] % 2 == 0)
      c[st][i] = 2;
    else
      c[i][fi] = 2;
  cnt = 0;
  while (findpath()) incflow();
  if (cnt < n) {
    cout << "Impossible" << endl;
    return 0;
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j)
      if (f[i][j] > 0) {
        b[i].push_back(j);
        b[j].push_back(i);
      }
  }
  res = 0;
  memset(k, true, sizeof(k));
  for (int i = 1; i <= n; ++i)
    if (k[i]) res++, dfs(i);
  cout << res << endl;
  for (int i = 1; i <= res; ++i) {
    cout << v[i].size() << " ";
    for (auto j : v[i]) cout << j << " ";
    cout << endl;
  }
}