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;
const int inf = 1e9 + 7, N = 205;
int n, a[205], cnt1, cnt0;
vector<int> link[N], rem;
vector<vector<int> > ans;
bool used[N];
struct maxFlow {
  vector<int> gr[N];
  int c[N][N], f[N][N], d[N], now[N];
  void addEdge(int u, int v, int w) {
    gr[u].push_back(v);
    gr[v].push_back(u);
    c[u][v] = w;
  }
  bool bfs() {
    for (int i = 0; i <= n + 1; ++i) d[i] = inf, now[i] = 0;
    d[0] = 0;
    queue<int> q;
    q.push(0);
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      for (int v : gr[u]) {
        if (d[v] == inf && f[u][v] < c[u][v]) d[v] = d[u] + 1, q.push(v);
      }
    }
    return d[n + 1] != inf;
  }
  int dfs(int u, int flow) {
    if (u == n + 1 || !flow) return flow;
    for (int &i = now[u]; i < gr[u].size(); ++i) {
      int v = gr[u][i];
      if (d[v] != d[u] + 1 || f[u][v] >= c[u][v]) continue;
      int nextFlow = dfs(v, min(flow, c[u][v] - f[u][v]));
      if (nextFlow) {
        f[u][v] += nextFlow;
        f[v][u] -= nextFlow;
        return nextFlow;
      }
    }
    return 0;
  }
  int solve() {
    while (bfs()) {
      while (dfs(0, inf))
        ;
    }
    int ans = 0;
    for (int u : gr[0]) ans += f[0][u];
    return ans;
  }
} mf;
void dfs(int u, int p) {
  used[u] = 1;
  rem.push_back(u);
  for (int v : link[u])
    if (v != p && !used[v]) {
      dfs(v, u);
    }
}
bool check(int num) {
  for (int i = 2; i * i <= num; ++i) {
    if (num % i == 0) return 0;
  }
  return 1;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 1; i <= n; ++i) {
    for (int j = i + 1; j <= n; ++j) {
      if (check(a[i] + a[j])) {
        if (a[i] % 2 == 0)
          mf.addEdge(j, i, 1);
        else
          mf.addEdge(i, j, 1);
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (a[i] % 2)
      mf.addEdge(0, i, 2), cnt1++;
    else
      mf.addEdge(i, n + 1, 2), cnt0++;
  }
  for (int i = 0; i <= n + 1; ++i) {
    sort(mf.gr[i].begin(), mf.gr[i].end());
    mf.gr[i].erase(unique(mf.gr[i].begin(), mf.gr[i].end()), mf.gr[i].end());
  }
  if (mf.solve() != 2 * cnt1 || !cnt1 || mf.solve() != 2 * cnt0)
    return cout << "Impossible\n", 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      if (mf.f[i][j]) link[i].push_back(j), link[j].push_back(i);
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (!used[i]) {
      rem.clear();
      dfs(i, i);
      ans.push_back(rem);
    }
  }
  cout << ans.size() << '\n';
  for (int i = 0; i < ans.size(); ++i) {
    cout << ans[i].size() << ' ';
    for (int j : ans[i]) cout << j << ' ';
    cout << '\n';
  }
  return 0;
}