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 double PI = acos(-1);
const double eps = 1e-9;
const int inf = 2000000000;
const long long infLL = 9000000000000000000;
inline int in() {
  int x;
  scanf("%d", &x);
  return x;
}
inline long long inl() {
  long long x;
  scanf("%I64d", &x);
  return x;
}
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int power(int x, int n) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0) {
    int y = power(x, n / 2);
    return y * y;
  } else
    return x * power(x, n - 1);
}
const int maxn = 5001;
const int maxm = 21111;
int node, src, des, edge;
int ver[maxm], flow[maxm], nxt[maxm];
int head[maxn], work[maxn], dis[maxn], q[maxn];
void init(int nodee, int srcc, int dest) {
  node = nodee;
  src = srcc;
  des = dest;
  for (int i = 0; i < node; i++) head[i] = -1;
  edge = 0;
}
void add_edge(int frm, int to, int flw) {
  ver[edge] = to, flow[edge] = flw, nxt[edge] = head[frm], head[frm] = edge++;
  ver[edge] = frm, flow[edge] = 0, nxt[edge] = head[to], head[to] = edge++;
}
bool bfs() {
  int i, u, v, l, r = 0;
  for (int i = 0; i < node; i++) dis[i] = -1;
  dis[q[r++] = src] = 0;
  for (l = 0; l < r; l++) {
    for (i = head[u = q[l]]; i >= 0; i = nxt[i])
      if (flow[i] && dis[v = ver[i]] < 0) {
        dis[q[r++] = v] = dis[u] + 1;
        if (v == des) return 1;
      }
  }
  return 0;
}
bool dfs(int u, int ex) {
  if (u == des) return ex;
  for (int &i = work[u], v, tmp; i >= 0; i = nxt[i]) {
    if (flow[i] && dis[v = ver[i]] == dis[u] + 1 &&
        (tmp = dfs(v, min(ex, flow[i]))) > 0) {
      flow[i] -= tmp;
      flow[i ^ 1] += tmp;
      return tmp;
    }
  }
  return 0;
}
int fast_dinic() {
  int i, res = 0, del;
  while (bfs()) {
    for (i = 0; i < node; i++) work[i] = head[i];
    while (del = dfs(src, inf)) res += del;
  }
  return res;
}
int a[205], prime[20010];
int n, cnt = 0, vis[205];
vector<int> ans[100];
void dfs2(int x, int pa, int ori) {
  vis[x] = 1;
  ans[cnt].push_back(x);
  for (int e = head[x]; e != -1; e = nxt[e]) {
    int v = ver[e];
    if (v == src || v == des || v == pa || vis[v]) continue;
    if (flow[e ^ ori] == 0) {
      dfs2(v, x, ori ^ 1);
    }
  }
}
int main() {
  n = in();
  init(n + 2, 0, n + 1);
  for (int i = 1; i <= n; i++) {
    a[i] = in();
    if (a[i] & 1) {
      add_edge(src, i, 2);
    } else {
      add_edge(i, des, 2);
    }
  }
  for (int i = 2; i < 150; i++) {
    if (prime[i] == 0) {
      for (int j = i * i; j < 20010; j += i) prime[j] = 1;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (a[i] & 1) {
      for (int j = 1; j <= n; j++) {
        if (a[j] & 1 && i == j) continue;
        if (prime[a[i] + a[j]] == 0) add_edge(i, j, 1);
      }
    }
  }
  int an = fast_dinic();
  if (an != n) {
    printf("Impossible\n");
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    if (a[i] & 1 && vis[i] == 0) {
      cnt++;
      dfs2(i, -1, 0);
    }
  }
  printf("%d\n", cnt);
  for (int i = 1; i <= cnt; i++) {
    printf("%d ", ans[i].size());
    for (int j = 0; j < ans[i].size(); j++) printf("%d ", ans[i][j]);
    printf("\n");
  }
  return 0;
}
