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("-O2")
using namespace std;
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << endl;
  err(++it, args...);
}
const int LIM = 1e5 + 5, MOD = 1e9 + 7;
int t, n, m, k, x, y;
int is_com[LIM];
vector<int> pr;
void sieve() {
  is_com[1] = 1;
  for (int i = 2; i < LIM; i++) {
    if (!is_com[i]) {
      pr.push_back(i);
      for (int j = i + i; j < LIM; j += i) {
        is_com[j] = 1;
      }
    }
  }
}
struct edge {
  int from, to, cap, flow, index;
  edge(int from, int to, int cap, int flow, int index)
      : from(from), to(to), cap(cap), flow(flow), index(index) {}
};
struct PushRelabel {
  int n;
  vector<vector<edge> > g;
  vector<long long> excess;
  vector<int> height, active, count;
  queue<int> Q;
  PushRelabel(int n)
      : n(n), g(n), excess(n), height(n), active(n), count(2 * n) {}
  void addEdge(int from, int to, int cap) {
    g[from].push_back(edge(from, to, cap, 0, g[to].size()));
    if (from == to) g[from].back().index++;
    g[to].push_back(edge(to, from, 0, 0, g[from].size() - 1));
  }
  void enqueue(int v) {
    if (!active[v] && excess[v] > 0) {
      active[v] = true;
      Q.push(v);
    }
  }
  void push(edge &e) {
    int amt = (int)min(excess[e.from], (long long)e.cap - e.flow);
    if (height[e.from] <= height[e.to] || amt == 0) return;
    e.flow += amt;
    g[e.to][e.index].flow -= amt;
    excess[e.to] += amt;
    excess[e.from] -= amt;
    enqueue(e.to);
  }
  void relabel(int v) {
    count[height[v]]--;
    int d = 2 * n;
    for (auto &it : g[v]) {
      if (it.cap - it.flow > 0) d = min(d, height[it.to] + 1);
    }
    height[v] = d;
    count[height[v]]++;
    enqueue(v);
  }
  void gap(int k) {
    for (int v = 0; v < n; v++) {
      if (height[v] < k) continue;
      count[height[v]]--;
      height[v] = max(height[v], n + 1);
      count[height[v]]++;
      enqueue(v);
    }
  }
  void discharge(int v) {
    for (int i = 0; excess[v] > 0 && i < g[v].size(); i++) push(g[v][i]);
    if (excess[v] > 0) {
      if (count[height[v]] == 1)
        gap(height[v]);
      else
        relabel(v);
    }
  }
  long long max_flow(int source, int dest) {
    count[0] = n - 1;
    count[n] = 1;
    height[source] = n;
    active[source] = active[dest] = 1;
    for (auto &it : g[source]) {
      excess[source] += it.cap;
      push(it);
    }
    while (!Q.empty()) {
      int v = Q.front();
      Q.pop();
      active[v] = false;
      discharge(v);
    }
    long long max_flow = 0;
    for (auto &e : g[source]) max_flow += e.flow;
    return max_flow;
  }
};
vector<vector<int> > v, ans;
vector<int> vs;
void dfs(int i, int c) {
  vs[i] = 1;
  ans[c].push_back(i + 1);
  for (auto &it : v[i]) {
    if (!vs[it]) dfs(it, c);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  sieve();
  cin >> n;
  vector<int> odd, even;
  vector<int> indodd, indeven;
  for (int i = 0; i < n; i++) {
    cin >> x;
    if (x & 1)
      odd.push_back(x), indodd.push_back(i);
    else
      even.push_back(x), indeven.push_back(i);
  }
  if (odd.size() != even.size()) {
    cout << "Impossible\n";
    return 0;
  }
  PushRelabel p(n + 2);
  for (int i = 0; i < even.size(); i++) {
    p.addEdge(odd.size() + 1 + i, n + 1, 2);
  }
  for (int i = 0; i < odd.size(); i++) {
    p.addEdge(0, i + 1, 2);
    for (int j = 0; j < even.size(); j++) {
      if (!is_com[odd[i] + even[j]]) {
        p.addEdge(i + 1, j + 1 + odd.size(), 1);
      }
    }
  }
  int ans1 = p.max_flow(0, n + 1);
  if (ans1 == n) {
    v.assign(n, vector<int>(0));
    for (int i = 0; i < odd.size(); i++) {
      for (auto &it : p.g[i + 1]) {
        if (it.flow == 1) {
          x = indodd[i];
          y = indeven[it.to - 1 - odd.size()];
          v[x].push_back(y);
          v[y].push_back(x);
        }
      }
    }
    vs.assign(n, 0);
    int c = 0;
    for (int i = 0; i < n; i++) {
      if (!vs[i]) ans.push_back(vector<int>(0)), dfs(i, c), c++;
    }
    cout << c << "\n";
    for (auto &it : ans) {
      cout << it.size() << " ";
      for (auto &it1 : it) {
        cout << it1 << " ";
      }
      cout << "\n";
    }
  } else
    cout << "Impossible\n";
  return 0;
}
