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>
const int maxn = 202;
int n, a[maxn], s, t, tot, pre[maxn], level[maxn], que[maxn], head, tail, flow,
    ans, out[maxn][maxn];
struct Edge {
  int nxt, v, cost;
} e[maxn + maxn * maxn << 1];
bool isprime(int x) {
  for (int i = 2, j = (int)sqrt(x); i <= j; ++i)
    if (x % i == 0) return 0;
  return 1;
}
void addedge(int u, int v, int c) {
  e[tot] = (Edge){pre[u], v, c};
  pre[u] = tot++;
  e[tot] = (Edge){pre[v], u, 0};
  pre[v] = tot++;
}
bool bfs() {
  head = tail = 1;
  memset(level, 0, sizeof level);
  level[s] = 1;
  que[tail++] = s;
  while (head < tail) {
    int u = que[head++];
    for (int it = pre[u]; it != -1; it = e[it].nxt)
      if (e[it].cost && !level[e[it].v]) {
        level[e[it].v] = level[u] + 1;
        que[tail++] = e[it].v;
      }
  }
  return level[t];
}
int min(int x, int y) { return x < y ? x : y; }
int dfs(int u, int lim) {
  if (u == t || !lim) return lim;
  int flow = 0, tmp;
  for (int it = pre[u]; it != -1; it = e[it].nxt)
    if (level[e[it].v] == level[u] + 1 && e[it].cost &&
        (tmp = dfs(e[it].v, min(e[it].cost, lim - flow)))) {
      e[it].cost -= tmp;
      e[it ^ 1].cost += tmp;
      flow += tmp;
    }
  return flow;
}
void print(int now, int flag) {
  level[now] = 1;
  out[ans][++out[ans][0]] = now;
  if (!flag)
    for (int it = pre[now]; it != -1; it = e[it].nxt)
      if (e[it].v != s && !level[e[it].v] && !e[it].cost) {
        print(e[it].v, flag ^ 1);
        return;
      }
  if (flag)
    for (int it = pre[now]; it != -1; it = e[it].nxt)
      if (e[it].v != t && !level[e[it].v] && e[it].cost) {
        print(e[it].v, flag ^ 1);
        return;
      }
}
int main() {
  scanf("%d", &n);
  s = 0, t = n + 1;
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  memset(pre, -1, sizeof pre);
  for (int i = 1; i <= n; ++i)
    if (a[i] & 1)
      addedge(s, i, 2);
    else
      addedge(i, t, 2);
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= n; ++j)
      if ((a[i] & 1) && !(a[j] & 1) && isprime(a[i] + a[j])) addedge(i, j, 1);
  while (bfs()) {
    int tmp;
    while ((tmp = dfs(s, 0x3f3f3f3f))) flow += tmp;
  }
  if (flow != n) {
    puts("Impossible");
    return 0;
  }
  memset(level, 0, sizeof level);
  for (int i = 1; i <= n; ++i)
    if ((a[i] & 1) && !level[i]) {
      print(i, 0);
      ++ans;
    }
  printf("%d\n", ans);
  for (int i = 0; i < ans; ++i)
    for (int j = 0; j <= out[i][0]; ++j)
      printf("%d%c", out[i][j], " \n"[j == out[i][0]]);
  return 0;
}
