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>
using namespace std;
struct Edge {
  int from, to, cap, flow;
};
const int MAXN = 210;
const int INF = 0x3f3f3f3f;
int a[MAXN], d[MAXN];
bool vis[MAXN];
vector<Edge> edges;
vector<int> g[MAXN];
void addEdge(int u, int v, int cap) {
  edges.push_back((Edge){u, v, cap, 0});
  g[u].push_back(edges.size() - 1);
  edges.push_back((Edge){v, u, 0, 0});
  g[v].push_back(edges.size() - 1);
}
bool bfs(int s, int t) {
  queue<int> q;
  q.push(s);
  memset(d, -1, sizeof(d));
  d[s] = 0;
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    for (int i = 0; i < g[x].size(); ++i) {
      Edge& e = edges[g[x][i]];
      if (d[e.to] == -1 && e.cap > e.flow) {
        d[e.to] = d[x] + 1;
        q.push(e.to);
        if (e.to == t) return true;
      }
    }
  }
  return false;
}
int dfs(int x, int t, int low) {
  if (x == t) return low;
  int tmp;
  for (int i = 0; i < g[x].size(); ++i) {
    Edge& e = edges[g[x][i]];
    if (d[e.to] == d[x] + 1 &&
        (tmp = dfs(e.to, t, min(low, e.cap - e.flow))) > 0) {
      e.flow += tmp;
      edges[g[x][i] ^ 1].flow -= tmp;
      return tmp;
    }
  }
  return 0;
}
int dinic(int s, int t) {
  int flow = 0;
  while (bfs(s, t)) {
    flow += dfs(s, t, INF);
  }
  return flow;
}
bool isPrime(int n) {
  for (int i = 2; i * i <= n; ++i) {
    if (n % i == 0) return false;
  }
  return true;
}
void findAns(int x, int flag, vector<int>& table) {
  vis[x] = true;
  table.push_back(x);
  for (int i = 0; i < g[x].size(); ++i) {
    Edge& e = edges[g[x][i]];
    if (!vis[e.to] && e.flow * flag > 0) {
      findAns(e.to, -flag, table);
      return;
    }
  }
}
int main() {
  int n;
  cin >> n;
  int oddCnt = 0;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    oddCnt += a[i] % 2;
  }
  bool possible = false;
  if (oddCnt == n - oddCnt) {
    int s = 0, t = n + 1;
    for (int i = 1; i <= n; ++i) {
      if (a[i] % 2 == 1) {
        addEdge(s, i, 2);
        for (int j = 1; j <= n; ++j) {
          if (a[j] % 2 == 0 && isPrime(a[i] + a[j])) {
            addEdge(i, j, 1);
          }
        }
      } else {
        addEdge(i, t, 2);
      }
    }
    if (dinic(s, t) == n) {
      possible = true;
      vector<vector<int>> ans;
      memset(vis, 0, sizeof(vis));
      vis[s] = vis[t] = 1;
      for (int i = 1; i <= n; ++i) {
        if (a[i] % 2 == 1 && !vis[i]) {
          vector<int> table;
          findAns(i, 1, table);
          ans.push_back(table);
        }
      }
      cout << ans.size() << endl;
      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];
        }
        cout << endl;
      }
    }
  }
  if (!possible) {
    cout << "Impossible" << endl;
  }
  return 0;
}
