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;
double EPS = 1e-9;
int INF = 2000000000;
long long INFF = 8000000000000000000LL;
double PI = acos(-1);
int dirx[8] = {-1, 0, 0, 1, -1, -1, 1, 1};
int diry[8] = {0, 1, -1, 0, -1, 1, -1, 1};
inline string IntToString(int a) {
  char x[100];
  sprintf(x, "%d", a);
  string s = x;
  return s;
}
inline int StringToInt(string a) {
  char x[100];
  int res;
  strcpy(x, a.c_str());
  sscanf(x, "%d", &res);
  return res;
}
inline string GetString(void) {
  char x[1000005];
  scanf("%s", x);
  string s = x;
  return s;
}
inline string uppercase(string s) {
  int n = (int)s.size();
  for (int(i) = (0); (i) < (n); ++(i))
    if (s[i] >= 'a' && s[i] <= 'z') s[i] = s[i] - 'a' + 'A';
  return s;
}
inline string lowercase(string s) {
  int n = (int)s.size();
  for (int(i) = (0); (i) < (n); ++(i))
    if (s[i] >= 'A' && s[i] <= 'Z') s[i] = s[i] - 'A' + 'a';
  return s;
}
inline void OPEN(string s) {
  freopen((s + ".in").c_str(), "r", stdin);
  freopen((s + ".out").c_str(), "w", stdout);
}
class MaxFlow {
 public:
  int n, start, end, cap[4005][4005];
  vector<int> adj[4005];
  void reset(int _n) {
    n = _n;
    start = 0;
    end = n - 1;
    for (int(i) = (0); (i) < (n); ++(i))
      for (int(j) = (0); (j) < (n); ++(j)) cap[i][j] = 0;
    for (int(i) = (0); (i) < (n); ++(i)) adj[i].clear();
  }
  void addEdge(int a, int b, int c) {
    adj[a].push_back(b);
    adj[b].push_back(a);
    cap[a][b] += c;
  }
  int bisa(void) {
    queue<pair<int, int> > q;
    int parent[n + 5];
    bool sudah[n + 5];
    memset(parent, -1, sizeof(parent));
    memset(sudah, 0, sizeof(sudah));
    q.push(make_pair(start, INF));
    parent[start] = -1;
    sudah[start] = 1;
    while (!q.empty()) {
      pair<int, int> now = q.front();
      q.pop();
      if (now.first == end) {
        int ix = now.first;
        while (ix != start) {
          cap[parent[ix]][ix] -= now.second;
          cap[ix][parent[ix]] += now.second;
          ix = parent[ix];
        }
        return now.second;
      }
      for (int(i) = (0); (i) < ((int)adj[now.first].size()); ++(i)) {
        int dest = adj[now.first][i];
        if (cap[now.first][dest] > 0) {
          int jalan = min(cap[now.first][dest], now.second);
          if (!sudah[dest]) {
            parent[dest] = now.first;
            sudah[dest] = 1;
            q.push(make_pair(dest, jalan));
          }
        }
      }
    }
    return 0;
  }
  int countMaxFlow(void) {
    int res = 0;
    while (1) {
      int x = bisa();
      if (x == 0) break;
      res += x;
    }
    return res;
  }
};
int n, x;
vector<int> genap, ganjil;
int data[205];
bool sudah[205][2];
MaxFlow MF;
vector<vector<int> > risan;
vector<int> ana;
void dfs(int now, int x) {
  if (sudah[now][x]) return;
  sudah[now][x] = 1;
  if (x == 0) {
    ana.push_back(ganjil[now]);
    for (int(i) = (0); (i) < ((int)genap.size()); ++(i)) {
      if (MF.cap[i + (int)ganjil.size() + 1][now + 1] > 0) dfs(i, 1);
    }
  } else {
    ana.push_back(genap[now]);
    for (int(i) = (0); (i) < ((int)ganjil.size()); ++(i)) {
      if (MF.cap[now + (int)ganjil.size() + 1][i + 1] > 0) dfs(i, 0);
    }
  }
}
bool pr(int x) {
  for (int i = 2; i * i <= x; ++i) {
    if (x % i == 0) return false;
  }
  return true;
}
int main() {
  scanf("%d", &n);
  for (int(i) = (0); (i) < (n); ++(i)) {
    scanf("%d", &data[i]);
    if (data[i] % 2 == 0)
      genap.push_back(i);
    else
      ganjil.push_back(i);
  }
  if ((int)ganjil.size() != (int)genap.size()) {
    puts("Impossible");
    return 0;
  }
  MF.reset(n + 2);
  for (int(i) = (0); (i) < ((int)ganjil.size()); ++(i)) MF.addEdge(0, i + 1, 2);
  for (int(i) = (0); (i) < ((int)genap.size()); ++(i))
    MF.addEdge((int)ganjil.size() + 1 + i, n + 1, 2);
  for (int(i) = (0); (i) < ((int)ganjil.size()); ++(i))
    for (int(j) = (0); (j) < ((int)genap.size()); ++(j))
      if (pr(data[ganjil[i]] + data[genap[j]])) {
        MF.addEdge(i + 1, (int)ganjil.size() + 1 + j, 1);
      }
  int mf = MF.countMaxFlow();
  if (mf != n) {
    puts("Impossible");
    return 0;
  }
  for (int(i) = (0); (i) < ((int)ganjil.size()); ++(i)) {
    ana.clear();
    dfs(i, 0);
    if ((int)ana.size() > 0) risan.push_back(ana);
  }
  printf("%d\n", (int)risan.size());
  for (int(i) = (0); (i) < ((int)risan.size()); ++(i)) {
    printf("%d", (int)risan[i].size());
    for (int(j) = (0); (j) < ((int)risan[i].size()); ++(j))
      printf(" %d", risan[i][j] + 1);
    puts("");
  }
  return 0;
}
