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;
int n, a[100001], dis[100001], cur[100001], s, t, col[100001], ccnt;
bool pr[100001], vis[100001];
vector<int> v[100001];
struct edge {
  int v, w, next;
  edge() {}
  edge(int _v, int _w, int _next) {
    v = _v;
    w = _w;
    next = _next;
  }
} e[100001 << 1];
int head[100001], head_[100001], sz;
void init() {
  memset(head, -1, sizeof(head));
  sz = 1;
}
void insert(int u, int v, int w, int *head_ = head) {
  e[++sz] = edge(v, w, head_[u]);
  head_[u] = sz;
}
bool bfs(int s) {
  queue<int> q;
  q.push(s);
  memset(dis, -1, sizeof(int) * (n + 2));
  dis[s] = 0;
  while (!q.empty()) {
    int p = q.front();
    q.pop();
    for (int i = head[p]; ~i; i = e[i].next) {
      if (~dis[e[i].v] || !e[i].w) continue;
      dis[e[i].v] = dis[p] + 1;
      q.push(e[i].v);
    }
  }
  return ~dis[t];
}
int dfs(int u, int lim) {
  if (u == t) return lim;
  if (!lim) return 0;
  int ans = 0;
  for (int i = cur[u]; ~i; i = e[i].next) {
    cur[u] = i;
    if (!e[i].w || dis[e[i].v] != dis[u] + 1) continue;
    int res = dfs(e[i].v, min(lim, e[i].w));
    e[i].w -= res;
    e[i ^ 1].w += res;
    lim -= res;
    ans += res;
    if (!lim) break;
  }
  return ans;
}
int dinic() {
  int ans = 0;
  while (bfs(s)) {
    for (int i = s; i <= t; ++i) cur[i] = head[i];
    ans += dfs(s, 0x3f3f3f3f);
  }
  return ans;
}
bool isprime(int x) {
  for (int i = 2; i * i <= x; ++i) {
    if (x % i == 0) return false;
  }
  return true;
}
void dfs2(int u, int c) {
  v[c].push_back(u);
  vis[u] = true;
  for (int i = head_[u]; ~i; i = e[i].next) {
    if (vis[e[i].v]) continue;
    dfs2(e[i].v, c);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  init();
  s = 0, t = n + 1;
  for (int i = 1; i <= 2e4; ++i) pr[i] = isprime(i);
  for (int i = 1; i <= n; ++i)
    for (int j = i + 1; j <= n; ++j)
      if (pr[a[i] + a[j]]) {
        if (a[i] & 1)
          insert(i, j, 1), insert(j, i, 0);
        else
          insert(j, i, 1), insert(i, j, 0);
      }
  for (int i = 1; i <= n; ++i)
    if (a[i] & 1)
      insert(s, i, 2), insert(i, s, 0);
    else
      insert(i, t, 2), insert(t, i, 0);
  int tmp = dinic();
  if (tmp != n) {
    printf("Impossible\n");
    return 0;
  }
  memset(head_, -1, sizeof(head_));
  for (int i = 1; i <= n; ++i) {
    for (int j = head[i]; ~j; j = e[j].next) {
      if (e[j].w || (j & 1) || e[j].v == t || e[j].v == s) continue;
      insert(i, e[j].v, 1, head_);
      insert(e[j].v, i, 1, head_);
    }
  }
  for (int i = 1; i <= n; ++i)
    if (!vis[i]) dfs2(i, ++ccnt);
  printf("%d\n", ccnt);
  for (int i = 1; i <= ccnt; ++i) {
    printf("%d ", v[i].size());
    for (int j = 0; j < v[i].size(); ++j) {
      printf("%d ", v[i][j]);
    }
    putchar(10);
  }
  return 0;
}