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 int maxn = 100086;
const int inf = 2.1e9;
const long long Inf = 999999999999999999;
const int mod = 1000000007;
const double eps = 1e-6;
const double pi = acos(-1);
int n, s, t;
int nums[206];
int vis[maxn], num[maxn];
int prime[maxn], cur;
bool check[maxn];
int Head[208], Next[maxn], v[maxn], w[maxn], cnt;
void init() {
  memset(Head, -1, sizeof(Head));
  cnt = 0;
  s = 0, t = n + 1;
}
void primes() {
  for (int i = 2; i < 20086; i++) {
    if (!check[i]) {
      prime[cur++] = i;
    }
    for (int j = 0; j < cur; j++) {
      if (i * prime[j] > 20086) {
        break;
      }
      check[i * prime[j]] = i;
      if (i % prime[j] == 0) {
        break;
      }
    }
  }
}
void add(int x, int y, int z) {
  if (x == y) {
    return;
  }
  v[cnt] = y;
  w[cnt] = z;
  Next[cnt] = Head[x];
  Head[x] = cnt++;
  v[cnt] = x;
  w[cnt] = 0;
  Next[cnt] = Head[y];
  Head[y] = cnt++;
}
bool bfs() {
  memset(vis, 0, sizeof(vis));
  for (int i = 0; i <= t; i++) {
    num[i] = Head[i];
  }
  vis[s] = 1;
  queue<int> q;
  q.push(s);
  int r = 0;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    int k = Head[u];
    while (k != -1) {
      if (!vis[v[k]] && w[k]) {
        vis[v[k]] = vis[u] + 1;
        q.push(v[k]);
      }
      k = Next[k];
    }
  }
  return vis[t];
}
int dfs(int u, int f) {
  if (u == t) {
    return f;
  }
  int &k = num[u];
  int sum = 0;
  while (k != -1) {
    if (vis[v[k]] == vis[u] + 1 && w[k]) {
      int d = dfs(v[k], min(f, w[k]));
      if (d > 0) {
        w[k] -= d;
        w[k ^ 1] += d;
        return d;
      }
    }
    k = Next[k];
  }
  return sum;
}
int Dinic() {
  int ans = 0;
  while (bfs()) {
    int f;
    while ((f = dfs(s, inf)) > 0) {
      ans += f;
    }
  }
  return ans;
}
stack<int> st;
void dfs(int t) {
  vis[t] = 1;
  st.push(t);
  for (int k = Head[t]; k != -1; k = Next[k]) {
    if (v[k] == 0 || v[k] == n + 1) {
      continue;
    }
    if (vis[v[k]]) {
      continue;
    }
    if (nums[t] % 2 == 0 && w[k]) {
      dfs(v[k]);
    }
    if (nums[t] % 2 == 1 && !w[k]) {
      dfs(v[k]);
    }
  }
  vis[t] = -1;
}
queue<int> q[208];
int main() {
  primes();
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &nums[i]);
  }
  init();
  for (int i = 1; i <= n; i++) {
    if (nums[i] & 1) {
      add(s, i, 2);
    } else {
      add(i, t, 2);
    }
    for (int j = i + 1; j <= n; j++) {
      if (!check[nums[i] + nums[j]]) {
        if (nums[i] & 1) {
          add(i, j, 1);
        } else {
          add(j, i, 1);
        }
      }
    }
  }
  if (Dinic() != n) {
    printf("Impossible\n");
    return 0;
  }
  int numi = 0;
  memset(vis, 0, sizeof(vis));
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      dfs(i);
      numi++;
      while (!st.empty()) {
        q[i].push(st.top());
        st.pop();
      }
    }
  }
  printf("%d\n", numi);
  for (int i = 1; i <= n; i++) {
    if (!q[i].empty()) {
      printf("%d", q[i].size());
      while (!q[i].empty()) {
        printf(" %d", q[i].front());
        q[i].pop();
      }
      printf("\n");
    }
  }
  return 0;
}
