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 = 1000;
const int inf = 0x3f3f3f3f;
int level[maxn], n, f, d;
list<int> g[100];
bool vis[300];
int num[300];
bool ok(int x) {
  for (int i = 2; i <= sqrt(x); ++i)
    if (x % i == 0) return false;
  return true;
}
struct node {
  int to, cap, rev;
  node(int _to, int _cap, int _rev) : to(_to), cap(_cap), rev(_rev) {}
};
vector<node> edge[maxn];
void add_node(int a, int b, int c) {
  edge[a].push_back(node(b, c, edge[b].size()));
  edge[b].push_back(node(a, 0, edge[a].size() - 1));
}
int bfs(int s, int t) {
  memset(level, -1, sizeof(level));
  queue<int> q;
  q.push(s);
  level[s] = 0;
  while (!q.empty()) {
    int tmp = q.front();
    q.pop();
    for (int i = 0; i < edge[tmp].size(); i++) {
      node &tmp1 = edge[tmp][i];
      if (level[tmp1.to] == -1 && tmp1.cap > 0) {
        level[tmp1.to] = level[tmp] + 1;
        q.push(tmp1.to);
      }
    }
  }
  return level[t] != -1;
}
int dfs(int s, int t, int flow) {
  if (s == t) return flow;
  int ans = 0;
  for (int i = 0; i < edge[s].size(); i++) {
    node &tmp = edge[s][i];
    if (flow > ans && tmp.cap > 0 && level[tmp.to] == level[s] + 1) {
      int d = dfs(tmp.to, t, min(tmp.cap, flow - ans));
      if (d > 0) {
        tmp.cap -= d;
        edge[tmp.to][tmp.rev].cap += d;
        ans += d;
      }
    }
  }
  if (ans == 0) level[s] = -1;
  return ans;
}
int dinic(int s, int t) {
  int ans = 0, tf;
  while (bfs(s, t))
    while (tf = dfs(s, t, inf)) ans += tf;
  return ans;
}
int t, s, a[300];
void dfs1(int u, int k) {
  if (vis[u]) return;
  g[k].push_back(num[u]);
  vis[u] = 1;
  for (node i : edge[u]) {
    if (i.to == t || i.to == s) continue;
    if (a[num[u]] & 1 && i.cap == 0 && !vis[i.to]) {
      dfs1(i.to, k);
    } else if (!(a[num[u]] & 1) && i.cap == 1 && !vis[i.to]) {
      dfs1(i.to, k);
    }
  }
}
int main() {
  int b, c;
  scanf("%d", &n);
  t = n + 2;
  s = n + 1;
  vector<int> jp[2];
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    jp[a[i] & 1].push_back(a[i]);
    if (a[i] & 1)
      num[jp[1].size() - 1] = i;
    else
      num[jp[0].size() - 1 + n / 2] = i;
  }
  if (jp[0].size() != jp[1].size()) {
    printf("Impossible\n");
    return 0;
  }
  int cnt = 0;
  for (int i = 0; i < jp[1].size(); i++) {
    for (int j = 0; j < jp[0].size(); j++) {
      if (ok(jp[1][i] + jp[0][j])) add_node(i, j + n / 2, 1);
    }
  }
  for (int i = 0; i < jp[0].size(); i++) {
    add_node(s, i, 2);
    add_node(i + n / 2, t, 2);
  }
  if (dinic(s, t) == n) {
    cnt = 0;
    for (int i = 0; i < n; i++) {
      if (vis[i]) continue;
      dfs1(i, cnt++);
    }
    cout << cnt << endl;
    for (int i = 0; i < cnt; i++) {
      printf("%d ", g[i].size());
      for (int j : g[i]) printf("%d ", j);
      cout << endl;
    }
  } else
    printf("Impossible\n");
  return 0;
}
