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;
using ll = long long;
using ld = long double;
using uint = unsigned int;
using ull = unsigned long long;
template <typename T>
using pair2 = pair<T, T>;
using pii = pair<int, int>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
ll myRand(ll B) { return (ull)rng() % B; }
clock_t startTime;
double getCurrentTime() {
  return (double)(clock() - startTime) / CLOCKS_PER_SEC;
}
const ll INF = (ll)1e18;
const ll C = (ll)1e15;
struct Dinic {
  struct Edge {
    int v, to;
    ll cap;
    Edge() : v(), to(), cap() {}
    Edge(int _v, int _to, ll _cap) : v(_v), to(_to), cap(_cap) {}
  };
  vector<Edge> ed;
  vector<vector<int>> g;
  int S, T;
  int n;
  vector<int> dist;
  vector<int> idx;
  Dinic() : ed(), g(), S(), T() {}
  Dinic(int _n, int _S, int _T) {
    n = _n;
    S = _S;
    T = _T;
    g.resize(n);
    for (int i = 0; i < n; i++) g[i].clear();
    ed = vector<Edge>();
  }
  void addEdge(int v, int to, ll cap) {
    g[v].push_back((int)ed.size());
    ed.push_back(Edge(v, to, cap));
    g[to].push_back((int)ed.size());
    ed.push_back(Edge(to, v, 0));
  }
  bool BFS() {
    dist.resize(n);
    vector<int> q;
    for (int i = 0; i < n; i++) dist[i] = n;
    q.push_back(S);
    dist[S] = 0;
    for (int i = 0; i < (int)q.size(); i++) {
      int v = q[i];
      for (int id : g[v]) {
        Edge e = ed[id];
        if (e.cap <= 0) continue;
        int u = e.to;
        if (dist[u] <= dist[v] + 1) continue;
        dist[u] = dist[v] + 1;
        q.push_back(u);
      }
    }
    return dist[T] < n;
  }
  ll dfs(int v, ll flow) {
    if (v == T || flow == 0) return flow;
    ll res = 0;
    for (int &i = idx[v]; i < (int)g[v].size(); i++) {
      int id = g[v][i];
      Edge e = ed[id];
      int to = e.to;
      if (dist[to] != dist[v] + 1) continue;
      ll df = dfs(to, min(flow, e.cap));
      flow -= df;
      res += df;
      ed[id].cap -= df;
      ed[id ^ 1].cap += df;
      if (flow == 0) return res;
    }
    return res;
  }
  ll Flow() {
    idx.resize(n);
    ll res = 0;
    while (BFS()) {
      for (int i = 0; i < n; i++) idx[i] = 0;
      res += dfs(S, INF);
    }
    return res;
  }
} G;
const int N = 20202;
bool pr[N];
int n, m;
pii a[222], b[222];
bool used[222];
vector<int> g[222];
int main() {
  startTime = clock();
  for (int i = 2; i < N; i++) pr[i] = 1;
  for (int i = 2; i < N; i++) {
    if (!pr[i]) continue;
    for (int j = 2 * i; j < N; j += i) pr[j] = false;
  }
  int k;
  scanf("%d", &k);
  for (int i = 1; i <= k; i++) {
    int x;
    scanf("%d", &x);
    if (x & 1)
      a[n++] = make_pair(x, i);
    else
      b[m++] = make_pair(x, i);
  }
  G = Dinic(n + m + 2, n + m, n + m + 1);
  for (int i = 0; i < n; i++) G.addEdge(n + m, i, 2);
  for (int i = 0; i < m; i++) G.addEdge(n + i, n + m + 1, 2);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++)
      if (pr[a[i].first + b[j].first]) G.addEdge(i, n + j, 1);
  int F = G.Flow();
  if (F < 2 * max(n, m)) {
    printf("Impossible\n");
    return 0;
  }
  for (int id = 2 * (n + m); id < (int)G.ed.size(); id += 2) {
    if (G.ed[id].cap == 0) {
      int v = G.ed[id].v, u = G.ed[id].to;
      g[v].push_back(u);
      g[u].push_back(v);
    }
  }
  n += m;
  for (int i = 0; i < n; i++) assert((int)g[i].size() == 2);
  vector<vector<int>> res;
  for (int i = 0; i < n; i++) {
    if (used[i]) continue;
    int x = i;
    int y = g[x][0];
    vector<int> cyc;
    while (!used[x]) {
      cyc.push_back(x);
      used[x] = 1;
      swap(x, y);
      y ^= g[x][0] ^ g[x][1];
    }
    res.push_back(cyc);
  }
  printf("%d\n", (int)res.size());
  for (auto w : res) {
    printf("%d", (int)w.size());
    for (int v : w) {
      if (v < n - m) {
        printf(" %d", a[v].second);
      } else {
        printf(" %d", b[v - (n - m)].second);
      }
    }
    printf("\n");
  }
  return 0;
}
