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 N = 205, INF = 0x3f3f3f3f;
struct Edge {
  int to, nxt, w;
} e[N * N * 2];
int n, a[N], S, T, head[N], cntE = 1, odd, even;
int tot, vis[N], d[N], mxfl;
vector<int> ans[N];
inline void add(int u, int v, int w) {
  e[++cntE] = (Edge){v, head[u], w};
  head[u] = cntE;
  e[++cntE] = (Edge){u, head[v], 0};
  head[v] = cntE;
}
bool ispri(int x) {
  for (int i = 2; i * i <= x; ++i)
    if (x % i == 0) return 0;
  return 1;
}
bool bfs() {
  queue<int> q;
  q.push(S);
  memset(d, 0, sizeof(d));
  d[S] = 1;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int i = head[u], v; i; i = e[i].nxt)
      if (e[i].w && !d[v = e[i].to]) {
        d[v] = d[u] + 1, q.push(v);
        if (v == T) return 1;
      }
  }
  return 0;
}
int dinic(int u, int flow) {
  if (u == T) return flow;
  int rest = flow, k;
  for (int i = head[u], v; i && rest; i = e[i].nxt)
    if (e[i].w && d[v = e[i].to] == d[u] + 1) {
      k = dinic(v, min(e[i].w, rest));
      if (!k) d[v] = 0;
      e[i].w -= k, e[i ^ 1].w += k, rest -= k;
    }
  return flow - rest;
}
void dfs(int u) {
  vis[u] = 1;
  ans[tot].push_back(u);
  for (int i = head[u]; i; i = e[i].nxt) {
    if (vis[e[i].to] || !e[i ^ (a[u] & 1)].w) continue;
    dfs(e[i].to);
    break;
  }
}
int main() {
  scanf("%d", &n);
  S = n + 1, T = S + 1;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
    if (a[i] & 1)
      add(S, i, 2), ++odd;
    else
      add(i, T, 2), ++even;
  }
  if (odd != even) return puts("Impossible"), 0;
  for (int i = head[S]; i; i = e[i].nxt)
    for (int j = head[T]; j; j = e[j].nxt)
      if (ispri(a[e[i].to] + a[e[j].to])) add(e[i].to, e[j].to, 1);
  while (bfs()) mxfl += dinic(S, INF);
  if (mxfl != n) return puts("Impossible"), 0;
  vis[S] = vis[T] = 1;
  for (int i = 1; i <= n; ++i)
    if ((a[i] & 1) && !vis[i]) ++tot, dfs(i);
  printf("%d\n", tot);
  for (int i = 1; i <= tot; ++i) {
    printf("%d ", ans[i].size());
    for (int j : ans[i]) printf("%d ", j);
    puts("");
  }
  return 0;
}
