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 = 210;
struct edge {
  int v, flow, next;
} e[N << 10];
int head[N], cnt;
void addedge(int u, int v, int flow) {
  e[cnt].v = v;
  e[cnt].next = head[u];
  e[cnt].flow = flow;
  head[u] = cnt++;
}
void add(int u, int v, int flow) {
  addedge(u, v, flow);
  addedge(v, u, 0);
}
bool check(int n) {
  for (int i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
int dis[N];
bool bfs(int s, int t) {
  queue<int> q;
  q.push(s);
  memset(dis, -1, sizeof(dis));
  dis[0] = 0;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int i = head[u]; i != -1; i = e[i].next) {
      if (e[i].flow && dis[e[i].v] == -1) {
        dis[e[i].v] = dis[u] + 1;
        q.push(e[i].v);
      }
    }
  }
  return dis[t] != -1;
}
int dfs(int u, int t, int flow) {
  if (u == t) return flow;
  int tot = flow;
  for (int i = head[u]; i != -1 && flow; i = e[i].next) {
    int x = e[i].v;
    if (dis[x] == dis[u] + 1 && e[i].flow > 0) {
      int v = dfs(x, t, min(flow, e[i].flow));
      e[i].flow -= v;
      e[i ^ 1].flow += v;
      flow -= v;
    }
  }
  return tot - flow;
}
int a[N];
vector<int> v[N];
bool vis[N];
vector<int> path[N];
int cc;
void dfs(int u) {
  vis[u] = true;
  path[cc].push_back(u);
  for (int i = 0; i < v[u].size(); i++)
    if (!vis[v[u][i]]) dfs(v[u][i]);
}
int main() {
  int n;
  scanf("%d", &n);
  int s = 0, t = n + 1;
  memset(head, -1, sizeof(head));
  int tot = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] % 2) {
      tot++;
      add(s, i, 2);
    } else
      add(i, t, 2);
  }
  for (int i = 1; i <= n; i++)
    if (a[i] % 2)
      for (int j = 1; j <= n; j++)
        if (a[j] % 2 == 0 && check(a[i] + a[j])) add(i, j, 1);
  if (tot * 2 != n) {
    printf("Impossible\n");
    return 0;
  }
  int ans = 0;
  while (bfs(s, t)) {
    ans += dfs(s, t, n * 2);
  }
  if (ans != n) {
    printf("Impossible\n");
    return 0;
  }
  for (int i = 1; i <= n; i++)
    if (a[i] % 2)
      for (int j = head[i]; j != -1; j = e[j].next)
        if (!e[j].flow) {
          v[i].push_back(e[j].v);
          v[e[j].v].push_back(i);
        }
  for (int i = 1; i <= n; i++)
    if (!vis[i]) {
      cc++;
      dfs(i);
    }
  cout << cc << endl;
  for (int i = 1; i <= cc; i++) {
    cout << path[i].size();
    for (int j = 0; j < path[i].size(); j++) cout << " " << path[i][j];
    cout << endl;
  }
  return 0;
}