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 N = 205;
const int M = N * N * 2;
const int INF = 1e9 + 5;
int a[N];
bool vis[N];
vector<int> v[N];
vector<int> ans[N];
struct edge {
  int u, v, cap, flow;
};
struct maxFlow {
  int fst[N], nxt[M], dis[N], cur[N], tot;
  edge a[M];
  queue<int> q;
  int S, T;
  void init(int _S, int _T) {
    S = _S, T = _T;
    memset(fst, -1, sizeof(fst));
    tot = 0;
  }
  void add(int x, int y, int c) {
    a[tot] = (edge){x, y, c, 0};
    nxt[tot] = fst[x];
    fst[x] = tot++;
  }
  void build(int x, int y, int c) {
    add(x, y, c);
    add(y, x, 0);
  }
  bool bfs() {
    int x, y;
    memset(dis, -1, sizeof(dis));
    dis[S] = 0;
    q.push(S);
    while (!q.empty()) {
      x = q.front();
      q.pop();
      for (int p = fst[x]; ~p; p = nxt[p]) {
        y = a[p].v;
        if (dis[y] == -1 && a[p].cap > a[p].flow) {
          dis[y] = dis[x] + 1;
          q.push(y);
        }
      }
    }
    return dis[T] != -1;
  }
  int dfs(int u, int pre) {
    int v, f, ret = 0;
    if (u == T || !pre) return pre;
    for (int &p = cur[u]; ~p; p = nxt[p]) {
      v = a[p].v;
      if (dis[v] == dis[u] + 1 &&
          (f = dfs(v, min(pre, a[p].cap - a[p].flow)))) {
        a[p].flow += f;
        a[p ^ 1].flow -= f;
        ret += f;
        pre -= f;
        if (!pre) break;
      }
    }
    return ret;
  }
  int solve() {
    int ret = 0;
    while (bfs()) {
      for (int i = 0; i < (T); i++) cur[i] = fst[i];
      ret += dfs(S, INF);
    }
    return ret;
  }
  void getAns(int n) {
    for (int i = 0; i < tot; i += 2) {
      if (a[i].cap != a[i].flow) continue;
      if (a[i].u >= 1 && a[i].u <= n && a[i].v >= 1 && a[i].v <= n) {
        v[a[i].u].push_back(a[i].v);
        v[a[i].v].push_back(a[i].u);
      }
    }
  }
} it;
bool Prime(int x) {
  for (int i = 2; i * i <= x; i++) {
    if (x % i == 0) return 0;
  }
  return 1;
}
void dfs(int x, int i) {
  if (vis[x]) return;
  vis[x] = 1;
  ans[i].push_back(x);
  for (int y : v[x]) dfs(y, i);
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  it.init(0, n + 1);
  for (int i = 1; i <= n; i++) {
    if (a[i] & 1) {
      it.build(it.S, i, 2);
      for (int j = 1; j <= n; j++) {
        if (Prime(a[i] + a[j])) it.build(i, j, 1);
      }
    } else
      it.build(i, it.T, 2);
  }
  if (it.solve() != n) {
    puts("Impossible");
  } else {
    it.getAns(n);
    int cnt = 0;
    for (int i = 1; i <= n; i++) {
      if (vis[i]) continue;
      dfs(i, cnt);
      cnt++;
    }
    printf("%d\n", cnt);
    for (int i = 0; i < (cnt); i++) {
      printf("%d", ans[i].size());
      for (int j = 0; j < (ans[i].size()); j++) printf(" %d", ans[i][j]);
      puts("");
    }
  }
  return 0;
}