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 inf = 1e9 + 7;
const int N = 210;
const int K = 2e4 + 10;
int n, a[N], p[K], pn, s, t;
bool b[K], vis[N];
struct G {
  int to[N][N];
  void clear() { memset(to, 0, sizeof(to)); }
  void add(int u, int v, int fl) { to[u][v] = fl; }
} g, r;
queue<int> q;
int dpt[N], cur[N], tmp, ans;
int cnt, prt[N][N];
void Init() {
  for (int i = 2; i < K; ++i) {
    if (!b[i]) p[++pn] = i;
    for (int j = 1; j <= pn; ++j) {
      if (i * p[j] >= K) break;
      b[i * p[j]] = 1;
      if (i % p[j] == 0) break;
    }
  }
  scanf("%d", &n);
  if (n & 1) {
    puts("Impossible");
    exit(0);
  }
  int cnt = 0;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    if (a[i] & 1) ++cnt;
  }
  if (cnt != (n >> 1)) {
    puts("Impossible");
    exit(0);
  }
  s = 0;
  t = n + 1;
  for (int i = 1; i <= n; ++i) {
    if (a[i] & 1)
      g.add(i, t, 2);
    else
      g.add(s, i, 2);
  }
  for (int i = 1; i <= n; ++i)
    if (!(a[i] & 1))
      for (int j = 1; j <= n; ++j)
        if (b[a[i] + a[j]] == 0) g.add(i, j, 1);
  r = g;
  ans = 0;
}
bool Bfs() {
  q = queue<int>();
  q.push(s);
  memset(dpt, 0, sizeof(dpt));
  dpt[s] = 1;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int v = s; v <= t; ++v)
      if (g.to[u][v] && !dpt[v]) {
        dpt[v] = dpt[u] + 1;
        if (v == t) return true;
        q.push(v);
      }
  }
  return false;
}
int Dfs(int u, int fl) {
  if (u == t) return fl;
  int rem = fl, tmp;
  for (int &v = cur[u]; v <= t && rem; ++v)
    if (g.to[u][v] && dpt[v] == dpt[u] + 1) {
      if (tmp = Dfs(v, min(rem, g.to[u][v]))) {
        rem -= tmp;
        g.to[u][v] -= tmp;
        g.to[v][u] += tmp;
      }
    }
  return fl - rem;
}
void Dfs1(int u) {
  vis[u] = 1;
  prt[cnt][++prt[cnt][0]] = u;
  for (int v = 1; v <= n; ++v)
    if (r.to[u][v] && !vis[v]) Dfs1(v);
}
void Print() {
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= n; ++j) r.to[i][j] ^= g.to[i][j];
  memset(vis, 0, sizeof(vis));
  cnt = 0;
  memset(prt, 0, sizeof(prt));
  for (int i = 1; i <= n; ++i)
    if (!vis[i]) {
      ++cnt;
      Dfs1(i);
    }
  printf("%d\n", cnt);
  for (int i = 1; i <= cnt; ++i) {
    for (int j = 0; j <= prt[i][0]; ++j) printf("%d ", prt[i][j]);
    puts("");
  }
}
int main() {
  Init();
  while (Bfs()) {
    memset(cur, 0, sizeof(cur));
    while (tmp = Dfs(s, inf)) ans += tmp;
  }
  if (ans != n)
    puts("Impossible");
  else
    Print();
  return 0;
}
