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;
namespace Sieve {
bool vis[100000 + 5];
int prime[100000 + 5];
int cnt;
void sieve(int n) {
  vis[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (!vis[i]) prime[++cnt] = i;
    for (int j = 1; j <= cnt && prime[j] * i <= n; j++) {
      vis[i * prime[j]] = 1;
      if (i % prime[j] == 0) break;
    }
  }
}
bool is_prime(int x) { return !vis[x]; }
}  // namespace Sieve
int n;
int a[100000 + 5];
struct edge {
  int from;
  int to;
  int next;
  int flow;
} E[100000 * 2 + 5];
int sz = 1;
int head[100000 + 5];
void add_edge(int u, int v, int w) {
  sz++;
  E[sz].from = u;
  E[sz].to = v;
  E[sz].next = head[u];
  E[sz].flow = w;
  head[u] = sz;
  sz++;
  E[sz].from = v;
  E[sz].to = u;
  E[sz].next = head[v];
  E[sz].flow = 0;
  head[v] = sz;
}
int deep[100000 + 5];
bool bfs(int s, int t) {
  for (int i = s; i <= t; i++) deep[i] = 0;
  queue<int> q;
  q.push(s);
  deep[s] = 1;
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    for (int i = head[x]; i; i = E[i].next) {
      int y = E[i].to;
      if (E[i].flow && !deep[y]) {
        deep[y] = deep[x] + 1;
        q.push(y);
      }
    }
  }
  return deep[t] > 0;
}
int dfs(int x, int t, int minf) {
  if (x == t) return minf;
  int rest = minf, k;
  for (int i = head[x]; i; i = E[i].next) {
    int y = E[i].to;
    if (E[i].flow && deep[y] == deep[x] + 1) {
      k = dfs(y, t, min(rest, E[i].flow));
      E[i].flow -= k;
      E[i ^ 1].flow += k;
      rest -= k;
      if (k == 0) deep[y] = 0;
      if (rest == 0) break;
    }
  }
  return minf - rest;
}
int dinic(int s, int t) {
  int ans = 0, now = 0;
  while (bfs(s, t)) {
    while ((now = dfs(s, t, 0x3f3f3f3f))) ans += now;
  }
  return ans;
}
int m;
bool vis[100000 + 5];
vector<int> res[100000 + 5];
void print(int x) {
  if (x >= 1 && x <= n) res[m].push_back(x);
  vis[x] = 1;
  for (int i = head[x]; i; i = E[i].next) {
    int y = E[i].to;
    if (vis[y]) continue;
    if ((a[x] % 2 == 0 && E[i ^ 1].flow == 0) ||
        (a[x] % 2 == 1 && E[i].flow == 0)) {
      print(y);
    }
  }
}
int main() {
  Sieve::sieve(20000);
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  int s = 0, t = n + 1;
  for (int i = 1; i <= n; i++) {
    if (a[i] % 2 == 1)
      add_edge(s, i, 2);
    else {
      add_edge(i, t, 2);
      continue;
    }
    for (int j = 1; j <= n; j++) {
      if (Sieve::is_prime(a[i] + a[j])) add_edge(i, j, 1);
    }
  }
  int flow = dinic(s, t);
  if (flow != n) {
    puts("Impossible");
  } else {
    m = 0;
    for (int i = 1; i <= n; i++) {
      if (!vis[i]) {
        m++;
        print(i);
      }
    }
    printf("%d\n", m);
    for (int i = 1; i <= m; i++) {
      printf("%d ", (int)res[i].size());
      for (int j = 0; j < (int)res[i].size(); j++) printf("%d ", res[i][j]);
      printf("\n");
    }
  }
}