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>
#pragma GCC optimize(3)
using namespace std;
int ne[30000], S, T, d[30000], q[30000], cnt;
int e[205], o[205], a[205], vis[205], qd;
deque<int> g[205];
struct node {
  int v, next, w;
} edge[30000];
bool ss(int x) {
  if (x < 2) return 0;
  for (int i = 2; i <= sqrt(x); i++)
    if (x % i == 0) return 0;
  return 1;
}
void add(int u, int v, int w) {
  edge[cnt].v = v;
  edge[cnt].next = ne[u];
  edge[cnt].w = w;
  ne[u] = cnt++;
  edge[cnt].v = u;
  edge[cnt].next = ne[v];
  edge[cnt].w = 0;
  ne[v] = cnt++;
}
bool bfs() {
  memset(d, -1, sizeof(d));
  d[S] = 0;
  int r = 0;
  q[r++] = S;
  for (int l = 0; l < r; l++) {
    int u = q[l];
    for (int i = ne[u]; i != -1; i = edge[i].next) {
      if (edge[i].w > 0 && d[edge[i].v] == -1) {
        d[edge[i].v] = d[u] + 1;
        q[r++] = edge[i].v;
      }
    }
  }
  return d[T] != -1;
}
int dfs(int u, int flow) {
  if (u == T) return flow;
  int re = 0, t;
  for (int i = ne[u]; i != -1; i = edge[i].next) {
    if (d[edge[i].v] == d[u] + 1 && edge[i].w > 0) {
      t = dfs(edge[i].v, min(edge[i].w, flow));
      if (!t) continue;
      edge[i].w -= t;
      edge[i ^ 1].w += t;
      flow -= t;
      re += t;
      if (!flow) break;
    }
  }
  if (re) return re;
  d[u] = -1;
  return 0;
}
void dfs2(int u, int co, int flag) {
  vis[u] = 1;
  if (flag)
    g[co].push_front(u);
  else
    g[co].push_back(u);
  for (int i = ne[u]; i != -1; i = edge[i].next) {
    if (edge[i].v == S || edge[i].v == T) continue;
    if ((i & 1) && edge[i].w == 1 && !vis[edge[i].v]) dfs2(edge[i].v, co, flag);
    if (!(i & 1) && edge[i].w == 0 && !vis[edge[i].v])
      dfs2(edge[i].v, co, abs(flag - (u == qd)));
  }
}
int main() {
  memset(ne, -1, sizeof(ne));
  int n, sum = 0, q = 0;
  cin >> n;
  S = 0, T = n + 1;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] & 1) {
      o[++o[0]] = i;
      add(S, i, 2);
    } else {
      e[++e[0]] = i;
      add(i, T, 2);
    }
  }
  if (o[0] != e[0]) {
    cout << "Impossible";
    return 0;
  }
  for (int i = 1; i <= o[0]; i++) {
    for (int j = 1; j <= e[0]; j++) {
      if (ss(a[o[i]] + a[e[j]])) {
        add(o[i], e[j], 1);
      }
    }
  }
  while (bfs()) {
    sum += dfs(S, 2147483647);
  }
  if (sum != n) {
    cout << "Impossible";
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    if (vis[i]) continue;
    qd = i;
    dfs2(i, ++q, 1);
  }
  cout << q << '\n';
  for (int i = 1; i <= q; i++) {
    cout << g[i].size() << ' ';
    while (!g[i].empty()) {
      cout << g[i].front() << ' ';
      g[i].pop_front();
    }
    cout << '\n';
  }
  return 0;
}
