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 = 200 + 10, maxm = 4e4 + 10, maxp = 2e4 + 10;
struct Graph {
  int ndn, edn, last[maxn];
  int u[maxm * 2], v[maxm * 2], c[maxm * 2], nxt[maxm * 2];
  void init(int _n) {
    ndn = _n;
    edn = 0;
    memset(last, -1, sizeof last);
  }
  void adde(int _u, int _v, int _c) {
    u[edn] = _u;
    v[edn] = _v;
    c[edn] = _c;
    nxt[edn] = last[_u];
    last[_u] = edn++;
  }
};
struct Dinic {
  Graph *G;
  int second, T, dist[maxn], cur[maxn];
  int bfs();
  int dfs(int, int);
  int solve(Graph *, int, int);
};
int N, A[maxn];
bool sieve[maxp];
bool vis[maxn];
vector<int> to[maxn];
vector<vector<int> > ans;
Graph G;
Dinic D;
void gen(int);
void dfs(int, vector<int> &);
int main() {
  for (int i = 2; i < maxp; i++)
    if (!sieve[i])
      for (int j = i * i; j < maxp; j += i) sieve[j] = 1;
  int second, T;
  while (~scanf("%d", &N)) {
    second = N + 1, T = N + 2;
    G.init(N + 2);
    for (int i = 1; i <= N; i++) scanf("%d", &A[i]);
    int sum = 0;
    for (int i = 1; i <= N; i++) {
      if (A[i] & 1)
        G.adde(second, i, 2), G.adde(i, second, 0);
      else
        G.adde(i, T, 2), G.adde(T, i, 0), sum += 2;
    }
    int eno = G.edn;
    for (int e0 = G.last[second]; ~e0; e0 = G.nxt[e0])
      for (int e1 = G.last[T]; ~e1; e1 = G.nxt[e1]) {
        int u = G.v[e0], v = G.v[e1];
        if (!sieve[A[u] + A[v]]) {
          G.adde(u, v, 1);
          G.adde(v, u, 0);
        }
      }
    vector<int> res;
    if (D.solve(&G, second, T) != sum)
      puts("Impossible");
    else {
      memset(vis, 0, sizeof vis);
      ans.clear();
      for (int i = 1; i <= N; i++) to[i].clear();
      for (int e = eno; e < G.edn; e += 2)
        if (!G.c[e]) {
          to[G.u[e]].push_back(G.v[e]);
          to[G.v[e]].push_back(G.u[e]);
        }
      for (int i = 1; i <= N; i++)
        if (!vis[i]) {
          res.clear();
          dfs(i, res);
          ans.push_back(res);
        }
      cout << ans.size() << endl;
      for (auto now : ans) {
        cout << now.size();
        for (auto ct : now) cout << " " << ct;
        puts("");
      }
    }
  }
  return 0;
}
void dfs(int u, vector<int> &ans) {
  vis[u] = 1;
  ans.push_back(u);
  for (auto v : to[u])
    if (!vis[v]) {
      dfs(v, ans);
      return;
    }
}
int Dinic::solve(Graph *g, int s, int t) {
  G = g;
  second = s;
  T = t;
  int res = 0;
  while (bfs()) {
    for (int i = 1; i <= G->ndn; i++) cur[i] = G->last[i];
    res += dfs(second, 1e9);
  }
  return res;
}
int Dinic::bfs() {
  memset(dist, -1, sizeof dist);
  dist[second] = 0;
  queue<int> que;
  que.push(second);
  while (que.size()) {
    int u = que.front();
    que.pop();
    for (int e = G->last[u]; ~e; e = G->nxt[e]) {
      int v = G->v[e];
      if (dist[v] == -1 && G->c[e] > 0) {
        dist[v] = dist[u] + 1;
        que.push(v);
      }
    }
  }
  return ~dist[T];
}
int Dinic::dfs(int u, int tmin) {
  if (u == T || tmin == 0) return tmin;
  int nflw = 0, f;
  for (int &e = cur[u]; ~e; e = G->nxt[e]) {
    int v = G->v[e];
    if (dist[u] + 1 == dist[v] && (f = dfs(v, min(tmin, G->c[e]))) > 0) {
      G->c[e] -= f;
      G->c[e ^ 1] += f;
      nflw += f;
      tmin -= f;
      if (tmin == 0) break;
    }
  }
  return nflw;
}
