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 = 1000000039;
const int maxn = 203;
int n;
int a[maxn];
bool is_prime(int x) {
  for (int i = 2; i * i <= x; ++i) {
    if (x % i == 0) return 0;
  }
  return 1;
}
struct edge {
  int to, c, f;
  edge(int _to = 0, int _c = 0, int _f = 0) { to = _to, c = _c, f = _f; }
  int rem() { return c - f; }
};
int m, s, t;
vector<int> g[2 * maxn];
vector<edge> edges;
void edge_add(int from, int to, int c) {
  g[from].push_back(int(edges.size()));
  edges.push_back(edge(to, c, 0));
  g[to].push_back(int(edges.size()));
  edges.push_back(edge(from, 0, 0));
}
void edge_upd(int idx, int f) {
  edges[idx].f += f;
  edges[idx ^ 1].f -= f;
}
vector<int> d;
bool bfs() {
  d.assign(2 * maxn, inf);
  queue<int> q;
  d[s] = 0;
  q.push(s);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int i : g[u]) {
      edge e = edges[i];
      if (e.rem() > 0 && d[e.to] == inf) {
        d[e.to] = d[u] + 1;
        q.push(e.to);
      }
    }
  }
  return d[t] != inf;
}
int p[2 * maxn];
int dfs(int u, int c_min) {
  if (u == t) return c_min;
  for (; p[u] < int(g[u].size()); ++p[u]) {
    edge e = edges[g[u][p[u]]];
    if (e.rem() > 0 && d[e.to] == d[u] + 1) {
      int f = dfs(e.to, min(e.rem(), c_min));
      if (f > 0) {
        edge_upd(g[u][p[u]], f);
        return f;
      }
    }
  }
  return 0;
}
vector<int> adj[maxn];
bool cycled[maxn];
int32_t main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    if (a[i] & 1)
      edge_add(0, i, 2);
    else
      edge_add(maxn + i, 2 * maxn - 1, 2);
    for (int j = 1; j < i; ++j) {
      if (is_prime(a[i] + a[j])) {
        if (a[i] & 1)
          edge_add(i, maxn + j, 1);
        else
          edge_add(j, maxn + i, 1);
      }
    }
  }
  int fl = 0;
  s = 0, t = 2 * maxn - 1;
  while (bfs()) {
    memset(p, 0, sizeof(p));
    int delta;
    do {
      delta = dfs(s, inf);
      fl += delta;
    } while (delta);
  }
  if (fl != n)
    printf("Impossible\n");
  else {
    for (int i = 1; i <= n; ++i) {
      if (a[i] & 1) {
        for (int idx : g[i]) {
          edge e = edges[idx];
          if (e.rem() == 0) adj[i].push_back(e.to - maxn);
        }
      } else {
        for (int idx : g[i + maxn]) {
          edge e = edges[idx];
          if (e.rem() == 1) adj[i].push_back(e.to);
        }
      }
    }
    vector<vector<int> > cycles;
    for (int i = 1; i <= n; ++i) {
      if (!cycled[i] && bool(a[i] & 1)) {
        vector<int> cycle;
        int prv = -1, cur = i;
        do {
          cycled[cur] = 1;
          cycle.push_back(cur);
          int old = cur;
          if (adj[cur][0] == prv)
            cur = adj[cur][1];
          else
            cur = adj[cur][0];
          prv = old;
        } while (cur != i);
        cycles.push_back(cycle);
      }
    }
    printf("%d\n", int(cycles.size()));
    for (size_t i = 0; i < cycles.size(); ++i) {
      printf("%d ", int(cycles[i].size()));
      for (size_t j = 0; j < cycles[i].size(); ++j) printf("%d ", cycles[i][j]);
      printf("\n");
    }
  }
}
