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;
int n, a[205];
int c[205][205], f[205][205];
int mf, s, t;
int cl[205], q[205], pre[205];
int fac[20004], np, pr[20004];
int deg[205], slt, sl[205];
vector<int> g[205];
vector<int> lt[205];
bool findp() {
  for (int i = 1; i <= t; ++i) cl[i] = 0;
  int L = 1, R = 0;
  cl[s] = 1;
  q[++R] = s;
  pre[s] = 0;
  while (L <= R) {
    int u = q[L++];
    for (int v = 1; v <= t; ++v)
      if (f[u][v] < c[u][v] && cl[v] == 0) {
        cl[v] = 1;
        q[++R] = v;
        pre[v] = u;
        if (v == t) return 1;
      }
    for (int v = 1; v <= t; ++v)
      if (f[v][u] > 0 && cl[v] == 0) {
        cl[v] = 1;
        q[++R] = v;
        pre[v] = -u;
      }
  }
  return 0;
}
void incf() {
  int d = 1e9, v = t;
  while (pre[v]) {
    int u = pre[v];
    if (u > 0)
      d = min(d, c[u][v] - f[u][v]);
    else {
      u = -u;
      d = min(d, f[v][u]);
    }
    v = u;
  }
  mf += d;
  v = t;
  while (pre[v]) {
    int u = pre[v];
    if (u > 0)
      f[u][v] += d;
    else {
      u = -u;
      f[v][u] -= d;
    }
    v = u;
  }
}
void sang(int n) {
  for (int i = 2; i <= n; ++i) {
    if (fac[i] == 0) {
      fac[i] = i;
      pr[++np] = i;
    }
    int j = 1;
    while (j <= np && pr[j] <= n / i) {
      fac[pr[j] * i] = pr[j];
      ++j;
    }
  }
}
void dfs(int u) {
  cl[u] = 1;
  lt[slt].push_back(u);
  sl[slt]++;
  for (int i = 0; i < deg[u]; ++i) {
    int v = g[u][i];
    if (cl[v] == 0) dfs(v);
  }
}
int main() {
  cin.tie(0);
  cout.tie(0);
  ios::sync_with_stdio(0);
  sang(20000);
  cin >> n;
  s = n + 1;
  t = n + 2;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    if (a[i] % 2)
      c[s][i] = 2;
    else
      c[i][t] = 2;
  }
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= n; ++j)
      if (a[i] % 2 && a[j] % 2 == 0 && fac[a[i] + a[j]] == a[i] + a[j])
        c[i][j] = 1;
  while (findp()) incf();
  if (mf < n) {
    cout << "Impossible";
    return 0;
  }
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= n; ++j)
      if (f[i][j]) {
        deg[i]++;
        g[i].push_back(j);
        deg[j]++;
        g[j].push_back(i);
      }
  memset(cl, 0, sizeof(cl));
  for (int i = 1; i <= n; ++i)
    if (cl[i] == 0) {
      ++slt;
      dfs(i);
    }
  cout << slt << "\n";
  for (int i = 1; i <= slt; ++i) {
    cout << sl[i] << " ";
    for (int j = 0; j < sl[i]; ++j) cout << lt[i][j] << " ";
    cout << "\n";
  }
}