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;
const int MAXV = 210;
const int MAXE = 40010;
const int inf = 1e9;
struct Edge {
  int u, v, cap, flow, next;
} grana[MAXE];
int n, m, E;
vector<int> odd, even, sol[MAXV], g[MAXV];
int head[MAXV], dist[MAXV];
int src, snk, k;
bool mark[MAXV];
int val[MAXV], a[MAXV];
bool is_prime(int x) {
  int rt = sqrt(x);
  for (int i = 2; i <= rt; i++)
    if (x % i == 0) return false;
  return true;
}
void add_edge(int u, int v, int c1, int c2) {
  grana[E++] = (Edge){u, v, c1, 0, head[u]};
  head[u] = E - 1;
  grana[E++] = (Edge){v, u, c2, 0, head[v]};
  head[v] = E - 1;
}
bool bfs() {
  fill(dist, dist + MAXV, -1);
  queue<int> q;
  dist[src] = 0;
  for (q.push(src); !q.empty(); q.pop()) {
    int u = q.front();
    for (int e = head[u]; e != -1; e = grana[e].next) {
      int v = grana[e].v;
      if (grana[e].cap > grana[e].flow && dist[v] == -1) {
        dist[v] = dist[u] + 1;
        q.push(v);
      }
    }
  }
  return (dist[snk] != -1);
}
int dfs(int u, int minCap) {
  if (minCap == 0) return 0;
  if (u == snk) return minCap;
  for (int e = head[u]; e != -1; e = grana[e].next) {
    int v = grana[e].v;
    if (dist[v] != dist[u] + 1) continue;
    int augment = dfs(v, min(minCap, grana[e].cap - grana[e].flow));
    if (augment) {
      grana[e].flow += augment;
      grana[e ^ 1].flow -= augment;
      return augment;
    }
  }
  return 0;
}
int Dinic() {
  int flow = 0, bf = 0;
  while (bfs()) {
    do {
      bf = dfs(src, inf);
      flow += bf;
    } while (bf);
  }
  return flow;
}
void fail() {
  printf("Impossible");
  exit(0);
}
void traversal(int u, int num) {
  mark[u] = true;
  sol[num].push_back(val[u]);
  for (int i = 0; i < 2; i++) {
    int v = g[u][i];
    if (!mark[v]) {
      traversal(v, num);
      return;
    }
  }
}
int main() {
  fill(head, head + MAXV, -1);
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    a[i] = x;
    if (x & 1)
      odd.push_back(x);
    else
      even.push_back(x);
  }
  if (odd.size() != even.size()) fail();
  src = n, snk = n + 1;
  for (int i = 0; i < even.size(); i++) {
    for (int j = 0; j < odd.size(); j++) {
      if (is_prime(even[i] + odd[j])) {
        add_edge(i, even.size() + j, 1, 0);
      }
    }
  }
  for (int i = 0; i < even.size(); i++) add_edge(src, i, 2, 0);
  for (int i = 0; i < odd.size(); i++) add_edge(even.size() + i, snk, 2, 0);
  int maxFlow = Dinic();
  if (maxFlow < n) fail();
  for (int i = 0; i < even.size(); i++) {
    for (int e = head[i]; e != -1; e = grana[e].next) {
      if (grana[e].flow == 1) {
        g[i].push_back(grana[e].v);
        g[grana[e].v].push_back(i);
      }
    }
  }
  int p = 0;
  for (int i = 0; i < even.size(); i++) {
    while (a[p] != even[i]) p++;
    val[i] = p + 1;
    p++;
  }
  p = 0;
  for (int i = 0; i < odd.size(); i++) {
    while (a[p] != odd[i]) p++;
    val[(int)(even.size()) + i] = p + 1;
    p++;
  }
  for (int i = 0; i < n; i++)
    if (!mark[i]) traversal(i, k++);
  printf("%d\n", k);
  for (int i = 0; i < k; i++) {
    printf("%d ", sol[i].size());
    for (auto xt : sol[i]) printf("%d ", xt);
    printf("\n");
  }
  return 0;
}
