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;
vector<vector<int> > re;
vector<pair<int, int> > e, o;
int n, c[404][404], sink, Find, pre[404], flow;
bool ch[100009], visit[404];
void init() {
  for (int i = 2; i * i <= 100000; i++) {
    if (!ch[i]) {
      for (int j = i * i; j <= 100000; j += i) ch[j] = 1;
    }
  }
}
void dfs(int cur) {
  visit[cur] = 1;
  if (cur == sink) Find = 1;
  if (Find) return;
  for (int i = 0; i <= sink && !Find; i++) {
    if (c[cur][i] > 0 && !visit[i]) {
      dfs(i);
      pre[i] = cur;
    }
  }
}
void solve() {
  init();
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int a;
    scanf("%d", &a);
    if (a % 2)
      o.push_back(pair<int, int>(a, i + 1));
    else
      e.push_back(pair<int, int>(a, i + 1));
  }
  for (int i = 0; i < o.size(); i++) {
    c[0][i + 1] = 2;
  }
  sink = e.size() + o.size() + 1;
  for (int i = 0; i < e.size(); i++) {
    c[o.size() + i + 1][sink] = 2;
  }
  for (int i = 0; i < o.size(); i++) {
    for (int j = 0; j < e.size(); j++) {
      int t = o[i].first + e[j].first;
      if (!ch[t]) c[i + 1][o.size() + j + 1] = 1;
    }
  }
  for (;;) {
    memset(visit, 0, sizeof(visit));
    Find = 0;
    dfs(0);
    if (!Find) break;
    flow++;
    int st, en;
    en = sink;
    while (en) {
      st = pre[en];
      c[st][en]--;
      c[en][st]++;
      en = st;
    }
  }
  if (flow != 2 * e.size()) {
    puts("Impossible");
    return;
  }
  memset(visit, 0, sizeof(visit));
  for (int i = 1; i < sink; i++) {
    if (visit[i]) continue;
    vector<int> tmp;
    int cur = i;
    while (!visit[cur]) {
      visit[cur] = 1;
      if (cur <= o.size())
        tmp.push_back(o[cur - 1].second);
      else
        tmp.push_back(e[cur - o.size() - 1].second);
      for (int i = 1; i < sink; i++) {
        if (!visit[i] && ((cur <= o.size() && c[i][cur] > 0) ||
                          cur > o.size() && c[cur][i] > 0)) {
          cur = i;
          break;
        }
      }
    }
    re.push_back(tmp);
  }
  printf("%d\n", re.size());
  for (int i = 0; i < re.size(); i++) {
    printf("%d ", re[i].size());
    for (int j = 0; j < re[i].size(); j++) printf("%d ", re[i][j]);
    puts("");
  }
}
int main(void) {
  solve();
  return 0;
}