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>
const int mod = 1000000007;
const int inf = 1000000009;
const long long INF = 1000000000000000009;
const long long big = 1000000000000000;
const long double eps = 1e-9;
using namespace std;
vector<int> P[1005];
int T[1005], tab[20005];
bool O[1005];
void sito(int n) {
  for (int i = 2; i * i <= n; i++)
    if (!tab[i])
      for (int j = i * i; j <= n; j += i) tab[j] = 1;
}
class DINIC {
 public:
  queue<int> Q;
  int n, source, sink;
  vector<int> cur, dis;
  vector<vector<int> > cap;
  vector<vector<int> > nei;
  bool bfs() {
    dis = vector<int>(n + 1, -1);
    dis[source] = 0;
    Q.push(source);
    while (!Q.empty()) {
      int u = Q.front();
      Q.pop();
      for (int v : nei[u])
        if (!~dis[v] && cap[u][v]) {
          dis[v] = dis[u] + 1;
          Q.push(v);
        }
    }
    return ~dis[sink];
  }
  int dfs(int u, int mincap) {
    if (!mincap || u == sink) return mincap;
    int res = 0;
    for (int i = cur[u]; i < nei[u].size(); i++) {
      if (!mincap)
        break;
      else
        cur[u] = i;
      int v = nei[u][i];
      if (dis[v] == dis[u] + 1) {
        int f = dfs(v, min(mincap, cap[u][v]));
        mincap -= f;
        res += f;
        cap[u][v] -= f;
        cap[v][u] += f;
      }
    }
    return res;
  }

 public:
  DINIC(int n, int source, int sink)
      : n(n),
        source(source),
        sink(sink),
        cap(n + 1, vector<int>(n + 1, 0)),
        nei(n + 1) {}
  int flow() {
    int res = 0;
    while (bfs()) {
      cur = vector<int>(n + 1, 0);
      res += dfs(source, 1000000001);
    }
    return res;
  }
  void addEdge(int fr, int to, int cap1, int cap2 = 0) {
    nei[fr].push_back(to);
    nei[to].push_back(fr);
    cap[fr][to] += cap1;
    cap[to][fr] += cap2;
  }
};
vector<vector<int> > W;
vector<int> A;
int main() {
  ios::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  tab[1] = 1;
  sito(20001);
  int n;
  cin >> n;
  DINIC D(n + 2, 0, n + 1);
  for (int i = 1; i <= n; i++) cin >> T[i];
  for (int i = 1; i <= n; i++)
    for (int j = i + 1; j <= n; j++) {
      if (!tab[T[i] + T[j]]) {
        if (T[i] % 2 == 0)
          D.addEdge(i, j, 1);
        else
          D.addEdge(j, i, 1);
      }
    }
  for (int i = 1; i <= n; i++) {
    if (T[i] % 2 == 0)
      D.addEdge(0, i, 2);
    else
      D.addEdge(i, n + 1, 2);
  }
  if (D.flow() != n) {
    cout << "Impossible";
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    if (!O[i] && T[i] % 2 == 0) {
      int x = i;
      while (!O[x]) {
        A.push_back(x);
        O[x] = 1;
        for (auto v : D.nei[x])
          if (!O[v] && ((D.cap[x][v] == 0 && T[x] % 2 == 0) ||
                        D.cap[v][x] == 0 && T[x] % 2 == 1)) {
            x = v;
            break;
          }
      }
      W.push_back(A);
      A.clear();
    }
  }
  cout << W.size() << "\n";
  for (int i = 0; i < W.size(); i++) {
    cout << W[i].size() << " ";
    for (auto v : W[i]) cout << v << " ";
    cout << "\n";
  }
  return 0;
}