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 double PI = acos(-1.0);
const int M = 300;
int S, T, n, m, p;
int gr[M][M];
bool vis[M];
int parent[M];
int cap[M][M];
int beg[M][M];
bool crivao[100000];
void build_crivo() {
  for (int i = 2; i < 100000; i++) {
    if (!crivao[i]) {
      for (int j = i + i; j < 100000; j += i) {
        crivao[j] = true;
      }
    }
  }
}
void insertEdge(int u, int v, int capacidade) {
  gr[u][v] = 1;
  cap[u][v] = capacidade;
  beg[u][v] = capacidade;
  gr[v][u] = 1;
  cap[v][u] = 0;
}
int bfs() {
  memset(vis, false, sizeof vis);
  memset(parent, -1, sizeof parent);
  vis[S] = true;
  queue<pair<int, int> > q;
  q.push(make_pair(S, INT_MAX));
  while (!q.empty()) {
    int u = q.front().first;
    int flow = q.front().second;
    q.pop();
    if (u == T) {
      for (int v = T; v != S; v = parent[v]) {
        cap[parent[v]][v] -= flow;
        cap[v][parent[v]] += flow;
      }
      return flow;
    }
    for (int v = 0; v < M; v++) {
      if (!gr[u][v]) continue;
      if (!vis[v] and cap[u][v] > 0) {
        q.push(make_pair(v, min(flow, cap[u][v])));
        vis[v] = true;
        parent[v] = u;
      }
    }
  }
  return 0;
}
int main(void) {
  build_crivo();
  S = 250;
  T = 251;
  scanf("%d", &n);
  vector<int> v;
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    v.push_back(x);
    if (x % 2) {
      insertEdge(i, T, 2);
    } else {
      insertEdge(S, i, 2);
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (!crivao[v[i] + v[j]]) {
        if (v[i] % 2) {
          insertEdge(j, i, 1);
        } else {
          insertEdge(i, j, 1);
        }
      }
    }
  }
  int ans = 0, flow;
  while (flow = bfs()) {
    ans += flow;
  }
  if (ans == n) {
    vector<vector<int> > resp;
    set<int> s;
    for (int i = 0; i < n; i++) {
      s.insert(i);
    }
    while (!s.empty()) {
      int k = *(s.begin());
      vector<int> x;
      while (s.count(k)) {
        x.push_back(k);
        s.erase(k);
        if (v[k] % 2) {
          for (int i = 0; i < n; i++) {
            if (beg[i][k] and !cap[i][k] and s.count(i)) {
              k = i;
              break;
            }
          }
        } else {
          for (int i = 0; i < n; i++) {
            if (beg[k][i] and !cap[k][i] and s.count(i)) {
              k = i;
              break;
            }
          }
        }
      }
      resp.push_back(x);
    }
    cout << resp.size() << endl;
    for (int i = 0; i < resp.size(); i++) {
      cout << resp[i].size() << " ";
      for (int j = 0; j < resp[i].size(); j++) {
        printf("%d ", resp[i][j] + 1);
      }
      cout << endl;
    }
  } else
    printf("Impossible\n");
  return 0;
}