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 maxn = 200 + 10;
const int INF = 1000000000;
struct Edge {
  int from, to, cap, flow;
};
bool operator<(const Edge& a, const Edge& b) {
  return a.from < b.from || (a.from == b.from && a.to < b.to);
}
struct ISAP {
  int n, m, s, t;
  vector<Edge> edges;
  vector<int> G[maxn];
  bool vis[maxn];
  int d[maxn];
  int cur[maxn];
  int p[maxn];
  int num[maxn];
  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(t);
    vis[t] = 1;
    d[t] = 0;
    while (!Q.empty()) {
      int x = Q.front();
      Q.pop();
      for (int i = 0; i < G[x].size(); i++) {
        Edge& e = edges[G[x][i] ^ 1];
        if (!vis[e.from] && e.cap > e.flow) {
          vis[e.from] = 1;
          d[e.from] = d[x] + 1;
          Q.push(e.from);
        }
      }
    }
    return vis[s];
  }
  void ClearAll(int n) {
    this->n = n;
    for (int i = 0; i < n; i++) G[i].clear();
    edges.clear();
  }
  void ClearFlow() {
    for (int i = 0; i < edges.size(); i++) edges[i].flow = 0;
  }
  int Augment() {
    int x = t, a = INF;
    while (x != s) {
      Edge& e = edges[p[x]];
      a = min(a, e.cap - e.flow);
      x = edges[p[x]].from;
    }
    x = t;
    while (x != s) {
      edges[p[x]].flow += a;
      edges[p[x] ^ 1].flow -= a;
      x = edges[p[x]].from;
    }
    return a;
  }
  int Maxflow(int s, int t) {
    this->s = s;
    this->t = t;
    int flow = 0;
    BFS();
    memset(num, 0, sizeof(num));
    for (int i = 0; i < n; i++) num[d[i]]++;
    int x = s;
    memset(cur, 0, sizeof(cur));
    while (d[s] < n) {
      if (x == t) {
        flow += Augment();
        x = s;
      }
      int ok = 0;
      for (int i = cur[x]; i < G[x].size(); i++) {
        Edge& e = edges[G[x][i]];
        if (e.cap > e.flow && d[x] == d[e.to] + 1) {
          ok = 1;
          p[e.to] = G[x][i];
          cur[x] = i;
          x = e.to;
          break;
        }
      }
      if (!ok) {
        int m = n - 1;
        for (int i = 0; i < G[x].size(); i++) {
          Edge& e = edges[G[x][i]];
          if (e.cap > e.flow) m = min(m, d[e.to]);
        }
        if (--num[d[x]] == 0) break;
        num[d[x] = m + 1]++;
        cur[x] = 0;
        if (x != s) x = edges[p[x]].from;
      }
    }
    return flow;
  }
  vector<int> Mincut() {
    BFS();
    vector<int> ans;
    for (int i = 0; i < edges.size(); i++) {
      Edge& e = edges[i];
      if (!vis[e.from] && vis[e.to] && e.cap > 0) ans.push_back(i);
    }
    return ans;
  }
  void Reduce() {
    for (int i = 0; i < edges.size(); i++) edges[i].cap -= edges[i].flow;
  }
  void print() {
    printf("Graph:\n");
    for (int i = 0; i < edges.size(); i++)
      printf("%d->%d, %d, %d\n", edges[i].from, edges[i].to, edges[i].cap,
             edges[i].flow);
  }
};
ISAP g;
vector<int> G[maxn];
bool vis[maxn];
int a[maxn];
bool is_prime[20005];
void pre() {
  memset(is_prime, true, sizeof(is_prime));
  is_prime[1] = false;
  for (int i = 2; i < 20005; i++)
    if (is_prime[i]) {
      for (int j = 2 * i; j < 20005; j += i) {
        is_prime[j] = false;
      }
    }
}
vector<int> ans[maxn];
void dfs(int u, int fa, int id) {
  ans[id].push_back(u);
  vis[u] = true;
  for (int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    if (vis[v] || v == fa) continue;
    dfs(v, u, id);
  }
}
int main() {
  int n;
  pre();
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
  }
  g.ClearAll(n + 2);
  int odd = 0, even = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] % 2) {
      odd++;
      g.AddEdge(0, i, 2);
    } else {
      even++;
      g.AddEdge(i, n + 1, 2);
    }
  }
  if (odd != even) {
    printf("Impossible\n");
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      if (is_prime[a[i] + a[j]]) {
        if (a[i] % 2) {
          g.AddEdge(i, j, 1);
        } else {
          g.AddEdge(j, i, 1);
        }
      }
    }
  }
  int f = g.Maxflow(0, n + 1);
  if (f == 2 * odd) {
    for (int i = 0; i < g.edges.size(); i++) {
      Edge e = g.edges[i];
      if (e.cap == 1 && e.flow == 1) {
        int u = e.from;
        int v = e.to;
        G[u].push_back(v);
        G[v].push_back(u);
      }
    }
  } else {
    printf("Impossible\n");
    return 0;
  }
  memset(vis, false, sizeof(vis));
  int now = 0;
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      dfs(i, -1, ++now);
    }
  }
  cout << now << endl;
  for (int i = 1; i <= now; i++) {
    cout << ans[i].size();
    for (int j = 0; j < ans[i].size(); j++) {
      cout << " " << ans[i][j] << " ";
    }
  }
  return 0;
}