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 S, T, n;
struct edge {
  int u, v, cap, next;
} edges[201000];
int head[500], nCount = 1;
void AddEdge(int U, int V, int C) {
  edges[++nCount].u = U;
  edges[nCount].v = V;
  edges[nCount].cap = C;
  edges[nCount].next = head[U];
  head[U] = nCount;
}
void add(int U, int V, int C) {
  AddEdge(U, V, C);
  AddEdge(V, U, 0);
}
int q[201000 * 2];
int layer[500];
bool CountLayer() {
  memset(layer, -1, sizeof(layer));
  int h = 0, t = 1;
  q[h] = S;
  layer[S] = 1;
  while (h < t) {
    int u = q[h++];
    for (int p = head[u]; p != -1; p = edges[p].next) {
      int v = edges[p].v;
      if (layer[v] == -1 && edges[p].cap) {
        layer[v] = layer[u] + 1;
        q[t++] = v;
      }
    }
  }
  return layer[T] != -1;
}
int DFS(int u, int flow) {
  int used = 0;
  if (u == T) return flow;
  for (int p = head[u]; p != -1; p = edges[p].next) {
    int v = edges[p].v;
    if (layer[v] == layer[u] + 1 && edges[p].cap) {
      int tmp = DFS(v, min(edges[p].cap, flow - used));
      edges[p].cap -= tmp;
      edges[p ^ 1].cap += tmp;
      used += tmp;
      if (used == flow) break;
    }
  }
  if (!used) layer[u] = -1;
  return used;
}
int Dinic() {
  int maxflow = 0;
  while (CountLayer()) maxflow += DFS(S, 0x3f3f3f3f);
  return maxflow;
}
bool isPrime[21000];
void getPrime() {
  memset(isPrime, true, sizeof(isPrime));
  isPrime[1] = false;
  isPrime[2] = true;
  for (int i = 2; i < 21000; i++) {
    if (!isPrime[i]) continue;
    for (int j = i + i; j < 21000; j += i) isPrime[j] = false;
  }
}
int a[500];
bool vis[500];
int sol[500][500];
void DFS2(int id, int u, int fa) {
  vis[u] = true;
  sol[id][++sol[id][0]] = u;
  for (int p = head[u]; p != -1; p = edges[p].next) {
    int v = edges[p].v;
    if (v == fa) continue;
    if (v == S || v == T) continue;
    if (vis[v]) continue;
    if (a[u] % 2) {
      if (!edges[p].cap) {
        edges[p].cap++;
        DFS2(id, v, u);
      }
    }
    if (!(a[u] % 2)) {
      if (edges[p].cap) {
        edges[p].cap--;
        DFS2(id, v, u);
      }
    }
  }
}
int main() {
  getPrime();
  memset(head, -1, sizeof(head));
  S = 500 - 2, T = 500 - 1;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) {
    if (a[i] % 2)
      add(S, i, 2);
    else
      add(i, T, 2);
  }
  for (int i = 1; i <= n; i++)
    for (int j = i + 1; j <= n; j++) {
      int x, y;
      if (a[i] % 2 && (!(a[j] % 2))) {
        x = i;
        y = j;
      } else if ((!(a[i] % 2)) && a[j] % 2) {
        x = j;
        y = i;
      } else
        continue;
      if (isPrime[a[x] + a[y]]) add(x, y, 1);
    }
  int maxflow = Dinic(), top = 0;
  if (maxflow == n) {
    for (int i = 1; i <= n; i++) {
      if (!vis[i]) {
        top++;
        DFS2(top, i, -1);
      }
    }
    printf("%d\n", top);
    for (int i = 1; i <= top; i++) {
      printf("%d", sol[i][0]);
      for (int j = 1; j <= sol[i][0]; j++) printf(" %d", sol[i][j]);
      printf("\n");
    }
  } else
    printf("Impossible\n");
  return 0;
}