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;
int n, m, res[210][210], vec[210], prv[210];
vector<int> adj[210];
const int s = 0;
const int t = 210 - 1;
int flow;
map<int, int> wh;
vector<vector<int> > ans;
vector<int> mat[210];
vector<bool> mark;
void init() {
  memset(res, 0, sizeof res);
  for (int i = 0; i < 210; i++) adj[i].clear();
}
bool isp(int x) {
  for (int i = 2; i * i <= x; i++)
    if (x % i == 0) return false;
  return true;
}
bool bfs() {
  queue<int> q;
  q.push(s);
  mark = vector<bool>(210, false);
  mark[s] = true;
  memset(prv, -1, sizeof prv);
  while (!q.empty()) {
    int u = q.front();
    if (u == t) return true;
    q.pop();
    for (int i = 0; i < adj[u].size(); ++i)
      if (!mark[adj[u][i]] && res[u][adj[u][i]] > 0) {
        int v = adj[u][i];
        mark[v] = true;
        prv[v] = u;
        q.push(v);
      }
  }
  return false;
}
void aug(int w, int minf) {
  if (prv[w] != -1) {
    aug(prv[w], min(minf, res[prv[w]][w]));
    res[prv[w]][w] -= flow;
    res[w][prv[w]] += flow;
  } else
    flow = minf;
}
int mf() {
  int mflow = 0;
  while (bfs()) {
    flow = 0;
    aug(t, 0x3f3f3f3f);
    mflow += flow;
  }
  return mflow;
}
void dfs(int u, int curr) {
  mark[u] = true;
  ans[curr].push_back(u);
  for (int i = 0; i < mat[u].size(); i++) {
    int v = mat[u][i];
    if (!mark[v]) {
      dfs(v, curr);
    }
  }
}
bool chk() {
  for (int i = 0; i < adj[t].size(); i++) {
    int v = adj[t][i];
    if (res[t][v] > 0)
      for (int j = 0; j < adj[v].size(); j++) {
        int w = adj[v][j];
        if (res[v][w]) {
          mat[v].push_back(w);
          mat[w].push_back(v);
        }
      }
  }
  mark = vector<bool>(210, false);
  int tab = 0;
  for (int i = 1; i <= n; i++)
    if (vec[i] % 2 == 0 && !mark[i]) {
      ans.push_back(vector<int>());
      dfs(i, tab++);
    }
  return true;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> vec[i];
    if (vec[i] % 2 == 0) {
      adj[s].push_back(i);
      adj[i].push_back(s);
      res[s][i] = 2;
    } else {
      res[i][t] = 2;
      adj[t].push_back(i);
      adj[i].push_back(t);
    }
  }
  for (int i = 1; i <= n; i++)
    if (vec[i] % 2 == 0)
      for (int j = 1; j <= n; j++)
        if (isp(vec[i] + vec[j])) {
          adj[i].push_back(j);
          adj[j].push_back(i);
          res[i][j] = 1;
        }
  int ff = mf();
  if (chk() && ff == n) {
    cout << ans.size() << "\n";
    for (int i = 0; i < ans.size(); i++) {
      cout << ans[i].size() << " ";
      for (int j = 0; j < ans[i].size(); j++)
        cout << ans[i][j] << " \n"[j == ans[i].size() - 1];
    }
  } else
    cout << "Impossible\n";
  return 0;
}