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 maxn = 1e5 + 5;
int n;
int a[maxn];
namespace Graph {
struct Node {
  int to;
  int next;
  int w;
} edge[maxn << 1];
int head[maxn], cnt = 1;
inline void add(int x, int y, int w) {
  edge[++cnt].next = head[x];
  edge[cnt].to = y;
  edge[cnt].w = w;
  head[x] = cnt;
}
}  // namespace Graph
using namespace Graph;
namespace NetworkFlow {
int s, t;
int dis[maxn];
int cur[maxn];
inline bool bfs() {
  queue<int> q;
  while (!q.empty()) q.pop();
  q.push(s);
  memset(dis, 0, sizeof(dis));
  memcpy(cur, head, sizeof(head));
  dis[s] = 1;
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    for (int i = head[x]; i != 0; i = edge[i].next) {
      int k = edge[i].to;
      int w = edge[i].w;
      if (!w) continue;
      if (dis[k]) continue;
      dis[k] = dis[x] + 1;
      q.push(k);
      if (k == t) return true;
    }
  }
  return false;
}
inline int Dinic(int x, int flow) {
  if (x == t) return flow;
  int rest = flow;
  for (int i = cur[x]; i != 0; i = edge[i].next) {
    cur[x] = i;
    int k = edge[i].to;
    int w = edge[i].w;
    if (dis[k] != dis[x] + 1) continue;
    if (!w) continue;
    int tmp = Dinic(k, min(rest, w));
    edge[i].w -= tmp;
    edge[i ^ 1].w += tmp;
    rest -= tmp;
  }
  return flow - rest;
}
inline int Maxflow() {
  int maxflow = 0;
  int flow = 0;
  while (bfs())
    if (flow = Dinic(s, 0x3f3f3f3f)) maxflow += flow;
  return maxflow;
}
}  // namespace NetworkFlow
using namespace NetworkFlow;
inline bool is_prime(int x) {
  if (x == 0 || x == 1) return false;
  for (int i = 2; i * i <= x; i++)
    if (x % i == 0) return false;
  return true;
}
int a1[maxn], n1;
int a2[maxn], n2;
vector<int> ans[maxn];
int v[maxn];
int res;
inline void dfs(int x) {
  v[x] = 1;
  ans[res].push_back(x);
  for (int i = head[x]; i != 0; i = edge[i].next) {
    int k = edge[i].to, w = edge[i].w;
    if (v[k]) continue;
    if (!edge[i ^ (a[x] & 1)].w) continue;
    dfs(k);
    break;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] & 1)
      a1[++n1] = i;
    else
      a2[++n2] = i;
  }
  if (n1 != n2) {
    printf("Impossible\n");
    return 0;
  }
  s = n + 1, t = n + 2;
  for (int i = 1; i <= n1; i++)
    for (int j = 1; j <= n2; j++)
      if (is_prime(a[a1[i]] + a[a2[j]]))
        add(a1[i], a2[j], 1), add(a2[j], a1[i], 0);
  for (int i = 1; i <= n1; i++) add(s, a1[i], 2), add(a1[i], s, 0);
  for (int i = 1; i <= n2; i++) add(a2[i], t, 2), add(t, a2[i], 0);
  if (Maxflow() != n) {
    printf("Impossible\n");
    return 0;
  }
  v[s] = v[t] = 1;
  for (int i = 1; i <= n1; i++)
    if (!v[a1[i]]) res++, dfs(a1[i]);
  printf("%d\n", res);
  for (int i = 1; i <= res; i++) {
    printf("%d ", ans[i].size());
    for (int j = 0; j < ans[i].size(); j++) printf("%d ", ans[i][j]);
    printf("\n");
  }
  return 0;
}
