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 = 2e2 + 4;
const int MAXP = 5e4;
const int MAXV = MAXN + 4;
const int inf = 1e9;
bool isp[MAXP];
int a[MAXN];
int n;
int c[MAXV][MAXV];
int f[MAXV][MAXV];
int d[MAXV];
int q[MAXV];
int ptr[MAXV];
int qh, qt;
const long long flow_inf = 1e18;
struct Dinic {
  struct FlowEdge {
    int v, u;
    long long cap, flow = 0;
    FlowEdge(int _v, int _u, long long _cap) : v(_v), u(_u), cap(_cap) {}
  };
  vector<FlowEdge> edges;
  vector<vector<int> > adj;
  int n, m = 0;
  int s, t;
  vector<int> level, ptr;
  vector<int> q;
  int qh, qt;
  void resize(int _n) {
    n = _n;
    q.resize(n);
    adj.resize(n);
    level.resize(n);
    ptr.resize(n);
  }
  Dinic() {}
  Dinic(int _n, int _s, int _t) {
    resize(_n);
    s = _s;
    t = _t;
  }
  void add_edge(int v, int u, long long cap) {
    edges.push_back(FlowEdge(v, u, cap));
    edges.push_back(FlowEdge(u, v, 0));
    adj[v].push_back(m);
    adj[u].push_back(m + 1);
    m += 2;
  }
  bool bfs() {
    while (qh < qt) {
      int v = q[qh++];
      for (int id : adj[v]) {
        if (edges[id].cap - edges[id].flow < 1) continue;
        if (level[edges[id].u] != -1) continue;
        level[edges[id].u] = level[v] + 1;
        q[qt++] = edges[id].u;
      }
    }
    return level[t] != -1;
  }
  long long dfs(int v, long long pushed = flow_inf) {
    if (pushed == 0) return 0;
    if (v == t) return pushed;
    for (int& cid = ptr[v]; cid < (int)adj[v].size(); cid++) {
      int id = adj[v][cid];
      int u = edges[id].u;
      if (level[v] + 1 != level[u] || edges[id].cap - edges[id].flow < 1)
        continue;
      long long tr = dfs(u, min(pushed, edges[id].cap - edges[id].flow));
      if (tr == 0) continue;
      edges[id].flow += tr;
      edges[id ^ 1].flow -= tr;
      return tr;
    }
    return 0;
  }
  long long flow() {
    long long f = 0;
    while (true) {
      ptr.assign(n, 0);
      level.assign(n, -1);
      level[s] = 0;
      qh = 0, qt = 1;
      q[0] = s;
      if (!bfs()) break;
      while (long long pushed = dfs(s)) {
        f += pushed;
      }
    }
    return f;
  }
};
bool used[MAXN];
int lst[MAXN];
int sz = 0;
int ancnt = 0;
void mark(int v) {
  used[v] = true;
  lst[sz++] = v + 1;
  for (auto i = 0; i != n; i++) {
    int w = (a[i] & 1) ? f[i][v] : f[v][i];
    if (w > 0 && !used[i]) mark(i);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  Dinic solver(MAXV, 0, 1);
  memset(c, 0, sizeof(c));
  memset(f, 0, sizeof(f));
  memset(isp, 0, sizeof(isp));
  isp[2] = false;
  for (auto i = 2; i != MAXP; i++) {
    if (isp[i]) continue;
    for (long long j = i * 1ll * i; j < MAXP; j += i) isp[j] = true;
  }
  scanf("%d", &n);
  for (auto i = 0; i != n; i++) scanf("%d", &a[i]);
  for (auto i = 0; i != n; i++)
    for (auto j = i + 1; j != n; j++) {
      if (isp[a[i] + a[j]]) continue;
      if (a[i] & 1)
        solver.add_edge(i + 2, j + 2, 1);
      else
        solver.add_edge(j + 2, i + 2, 1);
    }
  for (auto i = 0; i != n; i++) {
    if (a[i] & 1)
      solver.add_edge(0, i + 2, 2);
    else
      solver.add_edge(i + 2, 1, 2);
  }
  int flow = solver.flow();
  if (flow != n) {
    printf("Impossible");
    return 0;
  }
  for (auto e : solver.edges) {
    int v = e.v - 2, u = e.u - 2;
    if (v < 0) v += n + 2;
    if (u < 0) v += n + 2;
    f[v][u] = e.flow;
  }
  memset(used, 0, sizeof(used));
  for (auto i = 0; i != n; i++)
    if (!used[i]) {
      ancnt++;
      sz = 0;
      mark(i);
    }
  memset(used, 0, sizeof(used));
  printf("%d", ancnt);
  for (auto i = 0; i != n; i++)
    if (!used[i]) {
      printf("\n");
      sz = 0;
      mark(i);
      printf("%d ", sz);
      for (auto j = 0; j != sz; j++) printf("%d ", lst[j]);
    }
  return 0;
}
