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 int N = 222;
const int MaxE = 22222;
const int inf = 1e9;
int a[N], n;
struct Goldberg_Tarjan {
  int head[N], vis[N];
  vector<int> f[N], tmp;
  vector<vector<int> > ans;
  int cap[MaxE], to[MaxE], next[MaxE];
  int ecnt;
  int maxFlow;
  int height[N], C[N];
  int S, T, V;
  void init() { memset(head, -1, sizeof(head)); }
  void add(int s, int e, int c) {
    cap[ecnt] = c, to[ecnt] = e, next[ecnt] = head[s], head[s] = ecnt++;
    cap[ecnt] = 0, to[ecnt] = s, next[ecnt] = head[e], head[e] = ecnt++;
  }
  int ISAP(int x, int y) {
    int mn = V - 1, F = 0, R = y, t;
    if (x == T) return y;
    for (int i = head[x]; i != -1; i = next[i]) {
      if (cap[i] > 0) {
        if (height[x] == height[to[i]] + 1) {
          t = ISAP(to[i], min(cap[i], R));
          cap[i] -= t;
          cap[i ^ 1] += t;
          R -= t;
          F += t;
          if (height[S] >= V) return F;
          if (!R) break;
        }
        mn = min(mn, height[to[i]]);
      }
    }
    if (!F) {
      C[height[x]]--;
      if (!C[height[x]]) height[S] = V;
      height[x] = mn + 1;
      C[height[x]]++;
    }
    return F;
  }
  int GT(int src, int dst, int cnt) {
    S = src, T = dst, V = cnt;
    C[0] = V;
    maxFlow = 0;
    while (height[S] < V) maxFlow += ISAP(S, inf);
    return maxFlow;
  }
  void print() {
    for (int i = 1; i <= n; i++)
      if (a[i] & 1)
        for (int j = head[i]; j != -1 && to[j]; j = next[j])
          if (cap[j] == 0) f[i].push_back(to[j]), f[to[j]].push_back(i);
    for (int i = 1; i <= n; i++)
      if ((a[i] & 1) && !vis[i]) {
        int j = i, p = 0, k;
        tmp.clear();
        while (!vis[j]) {
          vis[j] = 1;
          tmp.push_back(j);
          for (k = 0; k < 2; k++)
            if (p != f[j][k]) break;
          p = j;
          j = f[j][k];
        }
        ans.push_back(tmp);
      }
    printf("%d\n", ans.size());
    for (int i = 0; i < ans.size(); i++) {
      printf("%d", ans[i].size());
      for (int j = 0; j < ans[i].size(); j++) printf(" %d", ans[i][j]);
      puts("");
    }
  }
} G;
bool ok(int x) {
  for (int i = 2; i * i <= x; i++)
    if (x % i == 0) return 0;
  return 1;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  G.init();
  for (int i = 1; i <= n; i++) {
    if (a[i] & 1) {
      G.add(0, i, 2);
      for (int j = 1; j <= n; j++)
        if (!(a[j] & 1) && ok(a[i] + a[j])) G.add(i, j, 1);
    } else
      G.add(i, n + 1, 2);
  }
  if (G.GT(0, n + 1, n + 2) != n)
    puts("Impossible");
  else
    G.print();
  return 0;
}