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 = 80005;
const int inf = 0x3f3f3f3f;
bool vis[maxn];
vector<int> odd, even;
vector<vector<int> > ans;
void init() {
  for (int i = 2; i < maxn; i++)
    if (!vis[i]) {
      for (int j = i * 2; j < maxn; j += i) vis[j] = 1;
    }
}
struct Edge {
  int from, to, cap, flow;
  Edge() {}
  Edge(int from, int to, int cap, int flow)
      : from(from), to(to), cap(cap), flow(flow) {}
};
struct Dinic {
  int n, m, s, t;
  vector<Edge> edges;
  vector<int> g[205];
  bool vis[205];
  int d[205];
  int cur[205];
  void init(int n) {
    this->n = n;
    edges.clear();
    for (int i = 0; i <= n; i++) g[i].clear();
  }
  void addedge(int from, int to, int cap) {
    edges.push_back(Edge(from, to, cap, 0));
    edges.push_back(Edge(to, from, 0, 0));
    m = edges.size();
    g[from].push_back(m - 2);
    g[to].push_back(m - 1);
  }
  bool bfs() {
    memset(vis, 0, sizeof vis);
    queue<int> q;
    q.push(s);
    d[s] = 0;
    vis[s] = 1;
    while (!q.empty()) {
      int x = q.front();
      q.pop();
      for (int i = 0; i < g[x].size(); i++) {
        Edge& e = edges[g[x][i]];
        if (!vis[e.to] && e.cap > e.flow) {
          vis[e.to] = 1;
          d[e.to] = d[x] + 1;
          q.push(e.to);
        }
      }
    }
    return vis[t];
  }
  int dfs(int x, int a) {
    if (x == t || a == 0) return a;
    int flow = 0, f;
    for (int& i = cur[x]; i < g[x].size(); i++) {
      Edge& e = edges[g[x][i]];
      if (d[x] + 1 == d[e.to] && (f = dfs(e.to, min(a, e.cap - e.flow))) > 0) {
        e.flow += f;
        edges[g[x][i] ^ 1].flow -= f;
        flow += f;
        a -= f;
        if (a == 0) break;
      }
    }
    return flow;
  }
  int maxflow(int s, int t) {
    this->s = s;
    this->t = t;
    int flow = 0;
    while (bfs()) {
      memset(cur, 0, sizeof cur);
      flow += dfs(s, inf);
    }
    return flow;
  }
} solve;
int main() {
  init();
  int n, a[222];
  while (scanf("%d", &n) > 0) {
    solve.init(n + 1);
    odd.clear();
    even.clear();
    for (int i = 1; i <= n; i++) {
      scanf("%d", a + i);
      if (a[i] % 2)
        odd.push_back(i);
      else
        even.push_back(i);
    }
    if (odd.size() != even.size()) {
      puts("Impossible");
      continue;
    }
    for (int i = 0; i < odd.size(); i++) solve.addedge(0, odd[i], 2);
    for (int i = 0; i < even.size(); i++) solve.addedge(even[i], n + 1, 2);
    for (int i = 0; i < odd.size(); i++)
      for (int j = 0; j < even.size(); j++)
        if (!vis[a[odd[i]] + a[even[j]]]) solve.addedge(odd[i], even[j], 1);
    int MAX = solve.maxflow(0, n + 1);
    if (MAX != n)
      puts("Impossible");
    else {
      ans.clear();
      vector<int> res;
      bool vis[222];
      memset(vis, 0, sizeof vis);
      for (int i = 1; i <= n; i++) {
        if (vis[i]) continue;
        res.clear();
        vis[i] = 1;
        int cnt = 0;
        res.push_back(i);
        int c = (a[i] % 2) ? 1 : -1;
        for (int j = 0; j < solve.g[i].size(); j++) {
          Edge x = solve.edges[solve.g[i][j]];
          if (x.flow == c) {
            res.push_back(x.to);
            vis[x.to] = 1;
            if (++cnt == 2) break;
          }
        }
        int now = res[res.size() - 1];
        while (now) {
          c *= -1;
          int newnow = 0;
          cnt = 0;
          for (int j = 0; j < solve.g[now].size(); j++) {
            Edge x = solve.edges[solve.g[now][j]];
            if (x.flow == c) {
              if (!vis[x.to]) {
                vis[x.to] = 1;
                newnow = x.to;
                res.push_back(x.to);
              }
              if (++cnt == 2) break;
            }
          }
          now = newnow;
        }
        ans.push_back(res);
      }
      cout << ans.size() << endl;
      for (int i = 0; i < ans.size(); i++) {
        vector<int> x = ans[i];
        cout << x.size();
        printf(" %d %d", x[1], x[0]);
        for (int j = 2; j < x.size(); j++) printf(" %d", x[j]);
        puts("");
      }
    }
  }
  return 0;
}
