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 nm = 302;
const int inf = 1000000000;
int n;
vector<int> adj[nm], adj2[nm];
int a[nm];
int cap[nm][nm], f[nm][nm];
int trace[nm];
vector<int> kq[nm];
int nkq;
bool fr[nm];
bool nt(int n) {
  if (n % 2 == 0) return 0;
  for (int i = 3; i * i <= n; i += 2)
    if (n % i == 0) return 0;
  return 1;
}
bool bfs() {
  memset(trace, 0, sizeof(trace));
  queue<int> q;
  q.push(1);
  trace[1] = -1;
  while (!q.empty()) {
    int i = q.front();
    q.pop();
    for (int k = 0; k < adj[i].size(); ++k) {
      int j = adj[i][k];
      if (!trace[j] && cap[i][j] - f[i][j] > 0) {
        trace[j] = i;
        if (j == n + 2) return 1;
        q.push(j);
      }
    }
    for (int k = 0; k < adj2[i].size(); ++k) {
      int j = adj2[i][k];
      if (!trace[j] && f[j][i] > 0) {
        trace[j] = -i;
        q.push(j);
      }
    }
  }
  return 0;
}
int tang() {
  int i, j, d = inf;
  j = n + 2;
  while (j != 1) {
    i = trace[j];
    if (i > 0)
      d = min(d, cap[i][j] - f[i][j]);
    else {
      i = -i;
      d = min(d, f[j][i]);
    }
    j = i;
  }
  j = n + 2;
  while (j != 1) {
    i = trace[j];
    if (i > 0)
      f[i][j] += d;
    else {
      i = -i;
      f[j][i] -= d;
    }
    j = i;
  }
  return d;
}
int maxflow() {
  int res = 0;
  while (bfs()) res += tang();
  return res;
}
void dfs(int i) {
  fr[i] = 0;
  kq[nkq].push_back(i - 1);
  if (a[i - 1] % 2 == 0) {
    for (int k = 0; k < adj[i].size(); ++k) {
      int j = adj[i][k];
      if (fr[j] && f[i][j] > 0) {
        dfs(j);
        return;
      }
    }
  } else {
    for (int k = 0; k < adj2[i].size(); ++k) {
      int j = adj2[i][k];
      if (fr[j] && f[j][i] > 0) {
        dfs(j);
        return;
      }
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= n; ++i) {
    if (a[i] % 2 == 0) {
      for (int j = 1; j <= n; ++j) {
        if (nt(a[i] + a[j])) {
          adj[1 + i].push_back(1 + j);
          adj2[1 + j].push_back(1 + i);
          cap[1 + i][1 + j] = 1;
        }
      }
    }
  }
  int s = 0;
  for (int i = 1; i <= n; ++i) {
    if (a[i] % 2 == 0) {
      adj[1].push_back(1 + i);
      cap[1][1 + i] = 2;
      s += 2;
    } else {
      adj[1 + i].push_back(n + 2);
      cap[1 + i][n + 2] = 2;
    }
  }
  int tmp = maxflow();
  if (tmp != s) {
    printf("Impossible\n");
    return 0;
  }
  memset(fr, 1, sizeof(fr));
  for (int i = 1; i <= n; ++i) {
    if (a[i] % 2 == 0 && fr[1 + i]) {
      nkq++;
      dfs(1 + i);
    }
  }
  printf("%d\n", nkq);
  for (int i = 1; i <= nkq; ++i) {
    printf("%d ", kq[i].size());
    for (int j = 0; j < kq[i].size(); ++j) printf("%d ", kq[i][j]);
    printf("\n");
  }
}