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, M = 500010;
int head[N], cur[N], dep[N], to[M], lim[M], nxt[M], vis[N], cnt = 1, n, m, s, t;
void add_edge(int u, int v, int l) {
  nxt[++cnt] = head[u], head[u] = cnt, lim[cnt] = l, to[cnt] = v;
  nxt[++cnt] = head[v], head[v] = cnt, lim[cnt] = 0, to[cnt] = u;
}
bool bfs() {
  queue<int> q;
  memset(dep, 0x3f, sizeof(dep));
  memset(vis, 0, sizeof(vis));
  memcpy(cur, head, sizeof(cur));
  dep[s] = 0;
  q.push(s);
  while (!q.empty()) {
    int h = q.front();
    vis[h] = 0;
    q.pop();
    for (int i = head[h]; i; i = nxt[i])
      if ((dep[to[i]] > dep[h] + 1) && lim[i]) {
        dep[to[i]] = dep[h] + 1;
        if (!vis[to[i]]) vis[to[i]] = 1, q.push(to[i]);
      }
  }
  if (dep[t] != 0x3f3f3f3f) return true;
  return false;
}
long long dfs(int rt = s, long long flow = 1e18) {
  if (rt == t) return flow;
  long long f = 0, ans = 0;
  for (int i = cur[rt]; i; i = nxt[i]) {
    cur[rt] = i;
    if (lim[i] && dep[to[i]] == dep[rt] + 1) {
      f = dfs(to[i], min(flow - ans, 1ll * lim[i]));
      if (f) {
        lim[i] -= f;
        lim[i ^ 1] += f;
        ans += f;
        if (ans == flow) break;
      }
    }
  }
  return ans;
}
long long dinic() {
  long long ans = 0, flow;
  while (bfs()) {
    while (flow = dfs()) ans += flow;
  }
  return ans;
}
int a[N], ans;
vector<int> od, ev, res[N];
bool isp(int x) {
  for (int i = 2; i * i <= x; ++i)
    if (x % i == 0) return 0;
  return 1;
}
void dfse(int);
void dfso(int);
void dfse(int rt) {
  vis[rt] = 1;
  res[ans].push_back(rt);
  for (int i = head[rt]; i; i = nxt[i])
    if (lim[i] && !vis[to[i]]) dfso(to[i]);
}
void dfso(int rt) {
  vis[rt] = 1;
  res[ans].push_back(rt);
  for (int i = head[rt]; i; i = nxt[i])
    if (!lim[i] && !vis[to[i]]) dfse(to[i]);
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    if (a[i] & 1)
      od.push_back(i);
    else
      ev.push_back(i);
  }
  if (od.size() != ev.size()) return puts("Impossible") & 0;
  s = n + 1, t = n + 2;
  for (auto i : od) add_edge(s, i, 2);
  for (auto i : ev) add_edge(i, t, 2);
  for (auto i : od)
    for (auto j : ev)
      if (isp(a[i] + a[j])) add_edge(i, j, 1);
  int f = dinic();
  if (f != n) return puts("Impossible") & 0;
  for (auto i : ev)
    if (!vis[i]) {
      ++ans;
      dfse(i);
    }
  cout << ans << endl;
  for (int i = 1; i <= ans; ++i) {
    cout << res[i].size() << " ";
    for (auto x : res[i]) cout << x << " ";
    cout << endl;
  }
}