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.
#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5;
int n, a[maxn + 5], ecnt;
int vis[maxn + 5], cnt, Ans[maxn + 5], dep[maxn + 5];
vector<vector<int> > ans;
struct Edge {
  int nxt, v, f;
} E[maxn + 5];
int h[maxn + 5], cur[maxn + 5];
void addedge(int u, int v, int f) {
  E[++ecnt].v = v, E[ecnt].f = f;
  E[ecnt].nxt = h[u], h[u] = ecnt;
  E[++ecnt].v = u, E[ecnt].f = 0;
  E[ecnt].nxt = h[v], h[v] = ecnt;
}
int isprime(int x) {
  for (int i = 2; i * i <= x; i++) {
    if (x % i == 0) return 0;
  }
  return 1;
}
int Q[maxn + 5];
int bfs() {
  int head = 1, tail = 0;
  for (int i = 0; i <= n + 1; i++) {
    cur[i] = h[i];
    dep[i] = -1;
  }
  Q[++tail] = 0;
  dep[0] = 1;
  while (head <= tail) {
    int u = Q[head++];
    for (int i = h[u]; i != -1; i = E[i].nxt) {
      int v = E[i].v, f = E[i].f;
      if (f && dep[v] == -1) {
        dep[v] = dep[u] + 1;
        if (v == n + 1) return 1;
        Q[++tail] = v;
      }
    }
  }
  return 0;
}
int dfs(int u, int lim) {
  if (lim == 0) return 0;
  if (u == n + 1) return lim;
  int ret = 0;
  for (int i = cur[u]; i != -1; i = E[i].nxt) {
    int v = E[i].v, f = E[i].f;
    if (dep[v] == dep[u] + 1 && f) {
      int t = dfs(v, min(lim, f));
      E[i].f -= t, E[i ^ 1].f += t;
      ret += t;
      lim -= t;
    }
  }
  if (ret == 0) {
    dep[u] = -1;
  }
  return ret;
}
int dinic() {
  int ans = 0;
  while (bfs()) {
    ans += dfs(0, 1e9);
  }
  return ans;
}
void Dfs(int u) {
  vis[u] = 1;
  Ans[++cnt] = u;
  for (int i = h[u]; i != -1; i = E[i].nxt) {
    int v = E[i].v;
    if (v == 0 or v == n + 1) continue;
    if (vis[v]) continue;
    if ((i % 2 == 0 && E[i].f == 0) or (i % 2 == 1 && E[i].f == 1)) Dfs(v);
  }
}
int main() {
  ecnt = 1;
  scanf("%d", &n);
  int S0 = 0, S1 = 0;
  for (int i = 0; i <= n + 1; i++) h[i] = -1;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] & 1)
      addedge(0, i, 2), S0 += 2;
    else
      addedge(i, n + 1, 2), S1 += 2;
  }
  if (S0 != S1) {
    puts("Impossible");
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (isprime(a[i] + a[j]) && (a[i] & 1)) {
        addedge(i, j, 1);
      }
    }
  }
  int T = dinic();
  if (T != S0) {
    puts("Impossible");
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    if (vis[i]) continue;
    cnt = 0;
    Dfs(i);
    vector<int> k;
    for (int i = 1; i <= cnt; i++) {
      k.push_back(Ans[i]);
    }
    ans.push_back(k);
  }
  printf("%d\n", ans.size());
  for (auto T : ans) {
    printf("%d ", T.size());
    for (auto t : T) printf("%d ", t);
    printf("\n");
  }
  return 0;
}
