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>
#pragma GCC optimize(2)
using namespace std;
const int inf = 0x3f3f3f3f;
const int N = 210, M = 1e5 + 10;
int n, a[N], vis[M], prime[M], sz;
vector<int> v[N];
int head[N], nex[M], to[M], w[M], tot = 1, s, t, h[N];
inline void ade(int a, int b, int c) {
  to[++tot] = b;
  nex[tot] = head[a];
  w[tot] = c;
  head[a] = tot;
}
inline void add(int a, int b, int c) {
  ade(a, b, c);
  ade(b, a, 0);
}
inline int bfs() {
  queue<int> q;
  q.push(s);
  memset(h, 0, sizeof h);
  h[s] = 1;
  while (q.size()) {
    int u = q.front();
    q.pop();
    for (int i = head[u]; i; i = nex[i]) {
      if (!h[to[i]] && w[i]) {
        h[to[i]] = h[u] + 1;
        q.push(to[i]);
      }
    }
  }
  return h[t];
}
int dfs(int x, int f) {
  if (x == t) return f;
  int fl = 0;
  for (int i = head[x]; i && f; i = nex[i]) {
    if (w[i] && h[to[i]] == h[x] + 1) {
      int mi = dfs(to[i], min(w[i], f));
      w[i] -= mi, w[i ^ 1] += mi, fl += mi, f -= mi;
    }
  }
  if (!fl) h[x] = -1;
  return fl;
}
inline int dinic() {
  int res = 0;
  while (bfs()) res += dfs(s, inf);
  return res;
}
inline void get_prime(int n) {
  for (int i = 2; i <= n; i++) {
    if (!prime[i]) {
      for (int j = i + i; j <= n; j += i) prime[j] = 1;
    }
  }
}
void dfs(int x, int fa, int flag, int k) {
  if (x == fa && flag) return;
  v[fa].push_back(x);
  vis[x] = 1;
  for (int i = head[x]; i; i = nex[i]) {
    if (!vis[to[i]] && w[i ^ k]) {
      dfs(to[i], fa, 1, k ^ 1);
      break;
    }
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin >> n;
  get_prime(M - 5);
  t = n + 1;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) {
    if (a[i] & 1)
      add(s, i, 2);
    else
      add(i, t, 2);
  }
  for (int i = 1; i <= n; i++) {
    if (!(a[i] & 1)) continue;
    for (int j = 1; j <= n; j++) {
      if (prime[a[i] + a[j]]) continue;
      add(i, j, 1);
    }
  }
  if (dinic() != n) {
    cout << "Impossible" << '\n';
    return 0;
  }
  vis[0] = vis[t] = 1;
  for (int i = 1; i <= n; i++) {
    if (!(a[i] & 1)) continue;
    if (vis[i]) continue;
    dfs(i, i, 0, 1);
    sz++;
  }
  cout << sz << '\n';
  for (int i = 1; i <= n; i++) {
    if (!v[i].size()) continue;
    cout << v[i].size();
    for (int j = 0; j < v[i].size(); j++) cout << ' ' << v[i][j];
    cout << '\n';
  }
  return 0;
}