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;
struct edge {
  int a, b, c;
  int next;
  edge() {}
  edge(int a, int b, int c, int n) : a(a), b(b), c(c), next(n) {}
} e[1010000 + 1000];
int p[70200], tol;
void add(int a, int b, int c) {
  e[tol] = edge(a, b, c, p[a]);
  p[a] = tol++;
}
int n, m;
int h[70005], gap[70005];
int INF;
int s, t;
int min(int a, int b) { return a < b ? a : b; }
int dfs(int u, int inf) {
  if (u == t) return inf;
  int d = inf, minh = INF;
  for (int i = p[u]; i != -1; i = e[i].next)
    if (e[i].c) {
      if (h[e[i].b] + 1 == h[u]) {
        int ret = dfs(e[i].b, min(d, e[i].c));
        e[i].c -= ret;
        e[i ^ 1].c += ret;
        d -= ret;
        if (h[s] >= INF) return inf - d;
        if (!d) break;
      }
      minh = min(minh, h[e[i].b]);
    }
  if (d == inf) {
    if (--gap[h[u]] == 0) h[s] = INF;
    h[u] = minh + 1;
    gap[h[u]]++;
  }
  return inf - d;
}
int maxflow() {
  int n = t + 1;
  INF = n;
  for (int i = 0; i < n; i++) h[i] = 0, gap[i] = 0;
  gap[0] = n;
  int mflow = 0;
  while (h[s] < INF) {
    int ret = dfs(s, 1 << 30);
    mflow += ret;
  }
  return mflow;
}
int a[1000];
int is[100000];
int v[100010];
int pr[20010];
void getprime() {
  memset(is, 0, sizeof(is));
  int cnt = 0;
  for (int i = 2; i < 20001; ++i) {
    if (!is[i]) pr[cnt++] = i;
    for (int j = 0; j < cnt && pr[j] * i < 20001; ++j) is[i * pr[j]] = 1;
  }
}
int isprime(int t) { return !is[t]; }
int pos;
vector<int> ans[10000];
void findans(int t) {
  ans[pos].push_back(t);
  v[t] = 1;
  for (int i = p[t]; i != -1; i = e[i].next) {
    if (v[e[i].b]) continue;
    if (a[t] & 1) {
      if (e[i].c) continue;
    } else {
      if (e[i ^ 1].c) continue;
    }
    findans(e[i].b);
  }
}
void solve() {
  memset(p, -1, sizeof(p));
  tol = 0;
  getprime();
  s = n, t = n + 1;
  int num = 0;
  for (int i = 0; i < n; ++i) {
    if (a[i] & 1) {
      ++num;
      add(s, i, 2);
      add(i, s, 0);
    } else {
      add(i, t, 2);
      add(t, i, 0);
    }
    for (int j = 0; j < i; ++j) {
      if (isprime(a[i] + a[j])) {
        if (a[i] & 1) {
          add(i, j, 1);
          add(j, i, 0);
        } else {
          add(j, i, 1);
          add(i, j, 0);
        }
      }
    }
  }
  pos = 0;
  int ret = maxflow();
  if (num * 2 != ret || num * 2 != n) {
    cout << "Impossible" << endl;
    return;
  } else {
    memset(v, 0, sizeof(v));
    for (int i = p[s]; i != -1; i = e[i].next) {
      if (v[e[i].b] == 0) {
        findans(e[i].b);
        pos++;
      }
    }
  }
  cout << pos << endl;
  for (int i = 0; i < pos; ++i) {
    cout << ans[i].size() << " ";
    for (int j = 0; j < ans[i].size(); ++j) printf("%d ", ans[i][j] + 1);
    cout << endl;
  }
}
int main() {
  cin >> n;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
  }
  solve();
}
