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;
struct edge {
  int from, to, c, f;
};
int n;
vector<int> arr;
vector<vector<int> > g;
vector<edge> e;
vector<int> cur;
vector<char> used;
bool prime(int a) {
  for (int i = 2; i <= sqrt(a); ++i) {
    if (a % i == 0) {
      return false;
    }
  }
  return true;
}
void add_edge(int a, int b, int cap) {
  edge e1 = {a, b, cap, 0}, e2 = {b, a, 0, 0};
  g[a].push_back(int(e.size()));
  e.push_back(e1);
  g[b].push_back(int(e.size()));
  e.push_back(e2);
}
vector<int> dp, ptr;
bool bfs() {
  dp.assign(n + 2, -1);
  dp[0] = 0;
  queue<int> q;
  q.push(0);
  while (!q.empty()) {
    int v = q.front();
    q.pop();
    for (int i = 0; i < g[v].size(); ++i) {
      int id = g[v][i], to = e[id].to;
      if (dp[to] == -1 && e[id].f < e[id].c) {
        dp[to] = dp[v] + 1;
        q.push(to);
      }
    }
  }
  return dp.back() != -1;
}
int dfs(int v, int flow) {
  if (flow == 0 || v == n + 1) {
    return flow;
  }
  for (; ptr[v] < g[v].size(); ++ptr[v]) {
    int id = g[v][ptr[v]], to = e[id].to;
    if (dp[to] == dp[v] + 1) {
      int pushed = dfs(to, min(flow, e[id].c - e[id].f));
      if (pushed > 0) {
        e[id].f += pushed;
        e[id ^ 1].f -= pushed;
        return pushed;
      }
    }
  }
  return 0;
}
void find_ans(int v) {
  cur.push_back(v);
  used[v] = true;
  for (int i = 0; i < g[v].size(); ++i) {
    int id = g[v][i], to = e[id].to;
    if (to >= 1 && to <= n && abs(e[id].f) == 1 && !used[to]) {
      find_ans(to);
    }
  }
}
int main() {
  scanf("%d", &n);
  arr.resize(n + 1);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &arr[i]);
  }
  g.resize(n + 2);
  for (int i = 1; i <= n; ++i) {
    if (arr[i] % 2 == 0) {
      add_edge(0, i, 2);
    } else {
      add_edge(i, n + 1, 2);
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = i + 1; j <= n; ++j) {
      if (prime(arr[i] + arr[j])) {
        if (arr[i] % 2 == 0) {
          add_edge(i, j, 1);
        } else {
          add_edge(j, i, 1);
        }
      }
    }
  }
  int max_flow = 0;
  for (;;) {
    if (!bfs()) {
      break;
    }
    ptr.assign(n + 2, 0);
    while (int pushed = dfs(0, 228)) {
      max_flow += pushed;
    }
  }
  vector<vector<int> > ans;
  if (max_flow == n) {
    used.resize(n + 2, false);
    for (int i = 1; i <= n; ++i) {
      if (!used[i]) {
        cur.clear();
        find_ans(i);
        ans.push_back(cur);
      }
    }
    printf("%d\n", int(ans.size()));
    for (int i = 0; i < ans.size(); ++i) {
      printf("%d ", int(ans[i].size()));
      for (int j = 0; j < ans[i].size(); ++j) {
        printf("%d ", ans[i][j]);
      }
      printf("\n");
    }
  } else {
    printf("Impossible");
  }
  return 0;
}