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 double eps = 1e-9;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const long long inf = (((long long)1) << 61) + 5;
const int maxn = 1005;
const int maxm = 255555;
struct Node {
  int to, next, cap;
} edge[maxm];
int tol;
int head[maxn];
int gap[maxn], dis[maxn], pre[maxn], cur[maxn];
void init() {
  tol = 0;
  memset(head, -1, sizeof head);
}
void add(int u, int v, int w, int rw = 0) {
  edge[tol].to = v;
  edge[tol].cap = w;
  edge[tol].next = head[u];
  head[u] = tol++;
  edge[tol].to = u;
  edge[tol].cap = rw;
  edge[tol].next = head[v];
  head[v] = tol++;
}
int sap(int start, int end, int N) {
  memset(dis, 0, sizeof(dis));
  memset(gap, 0, sizeof(gap));
  memcpy(cur, head, sizeof(head));
  int u = pre[start] = start, maxflow = 0, aug = -1;
  gap[0] = N;
  while (dis[start] < N) {
  loop:
    for (int &i = cur[u]; i != -1; i = edge[i].next) {
      int v = edge[i].to;
      if (edge[i].cap && dis[u] == dis[v] + 1) {
        if (aug == -1 || aug > edge[i].cap) aug = edge[i].cap;
        pre[v] = u;
        u = v;
        if (v == end) {
          maxflow += aug;
          for (u = pre[u]; v != start; v = u, u = pre[u]) {
            edge[cur[u]].cap -= aug;
            edge[cur[u] ^ 1].cap += aug;
          }
          aug = -1;
        }
        goto loop;
      }
    }
    int mindis = N;
    for (int i = head[u]; i != -1; i = edge[i].next) {
      int v = edge[i].to;
      if (edge[i].cap && mindis > dis[v]) {
        cur[u] = i;
        mindis = dis[v];
      }
    }
    if ((--gap[dis[u]]) == 0) break;
    gap[dis[u] = mindis + 1]++;
    u = pre[u];
  }
  return maxflow;
}
bool isprime(int x) {
  for (int i = 2; i * i <= x; i++)
    if (x % i == 0) return false;
  return true;
}
int a[205];
bool vis[205];
vector<int> ans[205];
void dfs(int u, vector<int> &g) {
  vis[u] = true;
  g.push_back(u);
  for (int i = head[u]; i != -1; i = edge[i].next) {
    int v = edge[i].to;
    if (vis[v]) continue;
    if (a[u] & 1) {
      if (edge[i].cap == 0) {
        dfs(v, g);
      }
    } else {
      if (edge[i].cap == 1) {
        dfs(v, g);
      }
    }
  }
}
int main() {
  int n;
  scanf("%d", &n);
  int s = 0, t = n + 1;
  init();
  int odd = 0, even = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] & 1)
      add(s, i, 2), odd++;
    else
      add(i, t, 2), even++;
  }
  if (odd != even) {
    puts("Impossible");
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    if (a[i] % 2 == 0) continue;
    for (int j = 1; j <= n; j++) {
      if (i == j) continue;
      if (a[j] & 1) continue;
      if (isprime(a[i] + a[j])) add(i, j, 1);
    }
  }
  if (sap(s, t, n + 2) != n) {
    puts("Impossible");
    return 0;
  }
  int res = 0;
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) dfs(i, ans[res++]);
  }
  printf("%d\n", res);
  for (int i = 0; i < res; i++) {
    printf("%d ", ans[i].size());
    for (int j = 0; j < ans[i].size(); j++) printf("%d ", ans[i][j]);
    puts("");
  }
  return 0;
}
