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;
int pri[41009], sp = 0;
bool ind[41009], is_pri[41009];
int n, a[41009], ans = 0, S, T, dis[41009], cur[41009];
int thd[41009], tto[41009], tnxt[41009], tcnt = 0, siz[41009];
int hd[41009], to[41009 << 1], nxt[41009 << 1], f[41009 << 1], cnt = 0;
struct DT {
  int u, va;
  friend bool operator<(DT x, DT y) { return x.va > y.va; }
};
priority_queue<DT> q;
int read();
void tadd(int u, int v) {
  tto[++tcnt] = v;
  tnxt[tcnt] = thd[u];
  thd[u] = tcnt;
  ++siz[u];
}
void add(int u, int v, int _f) {
  to[++cnt] = v;
  nxt[cnt] = hd[u];
  hd[u] = cnt;
  f[cnt] = _f;
}
bool dijk() {
  for (int i = 1; i <= T; ++i) dis[i] = 1047483647, cur[i] = hd[i];
  dis[S] = 0;
  q.push((DT){S, 0});
  while (q.size()) {
    DT now = q.top();
    q.pop();
    int u = now.u;
    if (dis[u] < now.va) continue;
    for (int i = hd[u], v; i; i = nxt[i])
      if (f[i] && (dis[v = to[i]] > dis[u] + 1)) {
        dis[v] = dis[u] + 1;
        q.push((DT){v, dis[v]});
      }
  }
  return dis[T] < 1047483647;
}
int dfs(int u, int flow) {
  if (!flow || u == T) return flow;
  int ret = 0;
  for (int i = cur[u], v; i; i = nxt[i])
    if (dis[v = to[i]] == dis[u] + 1) {
      int mb = flow < f[i] ? flow : f[i];
      int nowf = dfs(v, mb);
      f[i] -= nowf;
      ret += nowf;
      f[((i - 1) ^ 1) + 1] += nowf;
      flow -= nowf;
      cur[u] = i;
      if (!flow) return ret;
    }
  return ret;
}
void dfs2(int u, int ufa) {
  tadd(cnt, u);
  ind[u] = true;
  for (int i = hd[u], v; i; i = nxt[i])
    if ((f[i] ^ (a[u] & 1)) && (to[i] <= n) && (!ind[v = to[i]])) dfs2(v, u);
}
void get_pri() {
  for (int i = 2; i <= 3e4; ++i) {
    if (!is_pri[i]) pri[++sp] = i;
    for (int j = 1; j <= sp && 3e4 / i >= pri[j]; ++j)
      if (i % pri[j])
        is_pri[i * pri[j]] = true;
      else {
        is_pri[i * pri[j]] = true;
        break;
      }
  }
}
int main() {
  n = read();
  S = n + 1, T = n + 2;
  get_pri();
  for (int i = 1; i <= n; ++i) a[i] = read();
  for (int i = 1; i <= n; ++i)
    if (a[i] & 1)
      add(S, i, 2), add(i, S, 0);
    else
      add(i, T, 2), add(T, i, 0);
  for (int i = 1; i <= n; ++i)
    if (a[i] & 1)
      for (int j = 1; j <= n; ++j)
        if (!is_pri[a[i] + a[j]]) add(i, j, 1), add(j, i, 0);
  while (dijk()) ans += dfs(S, 1047483647);
  cnt = 0;
  if (ans < n) return puts("Impossible");
  for (int i = 1; i <= n; ++i)
    if (!ind[i]) ++cnt, dfs2(i, i);
  printf("%d\n", cnt);
  for (int i = 1; i <= cnt; ++i) {
    printf("%d ", siz[i]);
    for (int j = thd[i]; j; j = tnxt[j]) printf("%d ", tto[j]);
    puts("");
  }
  return 0;
}
int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c > '9' || c < '0') f = (c == '-') ? -1 : f, c = getchar();
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  return x * f;
}