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, K;
int arr[205];
int u[100005], v[100005], f[100005];
int lvl[205], ptr[205];
bool vis[205];
vector<int> graph[205];
vector<int> cgraph[205];
vector<vector<int>> vec;
void addedge(int a, int b, int k) {
  u[K] = a;
  v[K] = b;
  u[K ^ 1] = b;
  v[K ^ 1] = a;
  f[K] = k;
  graph[a].push_back(K);
  graph[b].push_back(K ^ 1);
  K += 2;
}
bool bfs() {
  queue<int> qu;
  qu.push(0);
  fill(lvl, lvl + N + 2, 0);
  lvl[0] = 1;
  while (qu.size()) {
    int n = qu.front();
    qu.pop();
    for (int e : graph[n]) {
      if (f[e] && !lvl[v[e]]) {
        lvl[v[e]] = lvl[u[e]] + 1;
        qu.push(v[e]);
      }
    }
  }
  return lvl[N + 1];
}
int dfs(int n, int p) {
  if (n == N + 1 || !p) {
    return p;
  }
  for (int &i = ptr[n]; i < graph[n].size(); i++) {
    int e = graph[n][i];
    if (f[e] && lvl[v[e]] > lvl[n]) {
      int k = dfs(v[e], min(f[e], p));
      if (!k) {
        continue;
      }
      f[e] -= k;
      f[e ^ 1] += k;
      return k;
    }
  }
  return 0;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> N;
  for (int i = 1; i <= N; i++) {
    cin >> arr[i];
    if (arr[i] % 2 == 1) {
      addedge(0, i, 2);
    } else {
      addedge(i, N + 1, 2);
    }
  }
  for (int i = 1; i <= N; i++) {
    for (int j = 1; j <= N; j++) {
      if (arr[i] % 2 == 1 && arr[j] % 2 == 0) {
        bool isp = 1;
        for (int k = 2; k <= sqrt(arr[i] + arr[j]); k++) {
          if ((arr[i] + arr[j]) % k == 0) {
            isp = 0;
            break;
          }
        }
        if (isp) {
          addedge(i, j, 1);
        }
      }
    }
  }
  int ans = 0;
  while (1) {
    if (!bfs()) {
      break;
    }
    fill(ptr, ptr + 2 + N, 0);
    int p;
    while ((p = dfs(0, INT_MAX / 2))) {
      ans += p;
    }
  }
  if (ans != N) {
    cout << "Impossible\n";
    return 0;
  }
  for (int e = 1; e < K; e += 2) {
    if (f[e] && u[e] != 0 && v[e] != 0 && u[e] != N + 1 && v[e] != N + 1) {
      cgraph[v[e]].push_back(u[e]);
      cgraph[u[e]].push_back(v[e]);
    }
  }
  for (int n = 1; n <= N; n++) {
    if (!vis[n]) {
      int c = n;
      vector<int> temp;
      while (!vis[c]) {
        temp.push_back(c);
        vis[c] = 1;
        for (int e : cgraph[c]) {
          if (!vis[e]) {
            c = e;
          }
        }
      }
      vec.push_back(temp);
    }
  }
  cout << vec.size() << "\n";
  for (auto k : vec) {
    cout << k.size() << " ";
    for (int n : k) {
      cout << n << " ";
    }
    cout << "\n";
  }
}