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>
using namespace std;
using LL = long long;
using PII = pair<int, int>;
const int N = 205;
int n, a[N], isprime[20005];
void init() {
  memset(isprime, -1, sizeof(isprime));
  for (int i = 2; i * i <= 20000; i++) {
    if (isprime[i])
      for (int j = i * i; j <= 20000; j += i) isprime[j] = 0;
  }
}
struct Edge {
  int u, v, cap, flow;
};
vector<Edge> es;
vector<int> g[N];
void addEdge(int u, int v, int cap) {
  es.push_back({u, v, cap, 0});
  es.push_back({v, u, 0, 0});
  g[u].push_back(es.size() - 2);
  g[v].push_back(es.size() - 1);
}
int iter[N], level[N];
bool bfs(int s, int t) {
  memset(level, -1, sizeof(level));
  level[s] = 0;
  queue<int> Q;
  Q.push(s);
  while (!Q.empty()) {
    int u = Q.front();
    Q.pop();
    for (auto i : g[u]) {
      Edge &e = es[i];
      if (level[e.v] < 0 && e.cap > e.flow) {
        level[e.v] = level[u] + 1;
        Q.push(e.v);
      }
    }
  }
  return level[t] > 0;
}
int dfs(int u, int t, int f) {
  if (u == t || f == 0) return f;
  int flow = 0, d;
  for (int &i = iter[u]; i < g[u].size(); i++) {
    Edge &e = es[g[u][i]];
    if (level[e.v] > level[u] &&
        (d = dfs(e.v, t, min(f, e.cap - e.flow))) > 0) {
      e.flow += d;
      es[g[u][i] ^ 1].flow -= d;
      flow += d;
      f -= d;
      if (f == 0) break;
    }
  }
  return flow;
}
int maxFlow(int s, int t) {
  int flow = 0;
  while (bfs(s, t)) {
    memset(iter, 0, sizeof(iter));
    flow += dfs(s, t, 1e9);
  }
  return flow;
}
vector<int> gg[N];
int main() {
  cin >> n;
  init();
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  int odd = 0, even = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] & 1)
      addEdge(0, i, 2), odd++;
    else
      addEdge(i, 201, 2), even++;
  }
  if (odd != even) return !printf("Impossible\n");
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      if (isprime[a[i] + a[j]] && (a[i] & 1)) {
        addEdge(i, j, 1);
      }
    }
  if (maxFlow(0, 201) != n) return !printf("Impossible\n");
  for (auto &e : es) {
    if (e.flow == 1) {
      gg[e.u].push_back(e.v);
      gg[e.v].push_back(e.u);
    }
  }
  vector<vector<int>> ans;
  int vis[N] = {0};
  for (int i = 1; i <= n; i++)
    if (!vis[i]) {
      int u = i, p = -1;
      vector<int> table;
      while (true) {
        if (vis[u]) break;
        vis[u] = 1;
        table.push_back(u);
        for (auto v : gg[u])
          if (v != p) {
            p = u, u = v;
            break;
          }
      }
      ans.push_back(table);
    }
  cout << ans.size() << endl;
  for (auto &vc : ans) {
    cout << vc.size();
    for (auto x : vc) cout << ' ' << x;
    cout << endl;
  }
}