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 = 5000 + 10, maxm = 60000 + 10;
const long long inf = 1e18;
int to[maxm];
long long cap[maxm];
int pr[maxm], last[maxn], pos[maxn], dis[maxn];
int cnt = 0;
void add(int fi, int se, int ca) {
  to[cnt] = se, cap[cnt] = ca, pr[cnt] = last[fi], last[fi] = cnt++;
  to[cnt] = fi, cap[cnt] = 0, pr[cnt] = last[se], last[se] = cnt++;
}
bool bfs(int source, int sink) {
  memset(dis, -1, sizeof dis);
  dis[source] = 0;
  queue<int> Q;
  Q.push(source);
  while (Q.size()) {
    int cur = Q.front();
    Q.pop();
    for (int i = last[cur]; i != -1; i = pr[i]) {
      if (cap[i] and dis[to[i]] == -1) {
        Q.push(to[i]);
        dis[to[i]] = dis[cur] + 1;
      }
    }
  }
  return dis[sink] != -1;
}
long long dfs(int source, int sink, long long untilNow) {
  if (source == sink) return untilNow;
  long long nowFlow = 0;
  for (int &i = pos[source]; i != -1; i = pr[i]) {
    if (cap[i] and dis[to[i]] == dis[source] + 1) {
      long long cur = dfs(to[i], sink, min(untilNow, cap[i]));
      cap[i] -= cur, cap[i ^ 1] += cur, untilNow -= cur, nowFlow += cur;
      if (untilNow == 0) return nowFlow;
    }
  }
  return nowFlow;
}
bool isp(int x) {
  for (int i = 2; i * i <= x; i++)
    if (x % i == 0) return false;
  return true;
}
int ans = 0;
vector<int> vans[1000];
bool markans[1000];
int a[1000];
void dfsans(int x, int mn, int mx) {
  vans[ans].push_back(x - 1);
  markans[x] = true;
  for (int i = last[x]; i != -1; i = pr[i]) {
    if (to[i] >= mn and to[i] <= mx and !markans[to[i]] &&
        ((a[x - 1] % 2 == 0 and cap[i] == 0) or
         (a[x - 1] % 2 == 1 and cap[i ^ 1] == 0))) {
      dfsans(to[i], mn, mx);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  memset(last, -1, sizeof last);
  int n;
  cin >> n;
  int sr = 1, si = n + 2;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] % 2 == 0)
      add(sr, i + 1, 2);
    else
      add(i + 1, si, 2);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (a[i] % 2 == 0 and isp(a[i] + a[j])) {
        add(i + 1, j + 1, 1);
      }
    }
  }
  long long maxFlow = 0;
  while (bfs(sr, si)) {
    for (int i = 0; i <= si; i++) pos[i] = last[i];
    maxFlow += dfs(sr, si, inf);
  }
  if (maxFlow != n) {
    cout << "Impossible" << endl;
  } else {
    for (int i = 2; i <= n + 1; i++) {
      if (!markans[i]) {
        ans++;
        dfsans(i, sr + 1, si - 1);
      }
    }
    cout << ans << endl;
    for (int i = 1; i <= ans; i++) {
      cout << vans[i].size() << " ";
      for (int j = 0; j < vans[i].size(); j++) cout << vans[i][j] << " ";
      cout << endl;
    }
  }
}