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;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pci = pair<char, int>;
using vi = vector<int>;
using vll = vector<ll>;
using vpii = vector<pii>;
const ll infll = 1e18 + 3;
const int maxn = 1e6 + 77;
const int nmax = 1e6 + 577;
const ll basell = 1e18 + 3;
const int mod = 1e9 + 7;
const ld eps = 1e-7;
const int inf = 1009000999;
const int nv = 100505;
const int baseint = 1000200013;
const ld PI = acos(-1.0);
inline bool EQ(ld a, ld b) { return fabs(a - b) < 1e-9; }
inline bool IF(int a, int b, int c) { return (a >= b && a <= c); }
inline bool IFS(int a, int b, int c) { return (a > b && a < c); }
struct edge {
  int a, b, cap, flow;
};
int n, s, t, d[2601], ptr[2601], q[2601];
vector<edge> e;
vector<int> g[2601];
void add_edge(int a, int b, int cap) {
  edge e1 = {a, b, cap, 0};
  edge 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);
}
bool bfs() {
  int qh = 0, qt = 0;
  q[qt++] = s;
  memset(d, -1, sizeof(d));
  d[s] = 0;
  while (qh < qt && d[t] == -1) {
    int v = q[qh++];
    for (int i = 0; i < ((int)(g[v]).size()); ++i) {
      int id = g[v][i], to = e[id].b;
      if (d[v] != -1 && d[to] == -1 && e[id].flow < e[id].cap) {
        q[qt++] = to;
        d[to] = d[v] + 1;
      }
    }
  }
  return d[t] != -1;
}
int dfs(int v, int flow) {
  if (!flow) return 0;
  if (v == t) return flow;
  for (; ptr[v] < ((int)(g[v]).size()); ++ptr[v]) {
    int id = g[v][ptr[v]], to = e[id].b;
    if (d[to] != d[v] + 1) continue;
    int push = dfs(to, min(flow, e[id].cap - e[id].flow));
    if (push) {
      e[id].flow += push;
      e[id ^ 1].flow -= push;
      return push;
    }
  }
  return 0;
}
int dinic() {
  int flow = 0;
  while (bfs()) {
    memset(ptr, 0, sizeof(ptr));
    while (int push = dfs(s, inf)) {
      flow += push;
    }
  }
  return flow;
}
int m;
int a[501];
vpii even, odd;
bool prime(int a) {
  for (int i = 2; i <= (int)sqrt(a); ++i) {
    if (a % i == 0) return false;
  }
  return true;
}
vi c[1001];
vi ans[1001];
int used[1001];
int cnt = 0;
void dfs(int u) {
  used[u] = 1;
  if (u <= m) {
    ans[cnt].push_back(even[u - 1].second + 1);
  } else {
    ans[cnt].push_back(odd[u - m - 1].second + 1);
  }
  for (int i = 0; i < ((int)(c[u]).size()); ++i) {
    int h = c[u][i];
    if (!used[h]) {
      dfs(h);
    }
  }
}
int inline solve() {
  ios::sync_with_stdio(NULL), cin.tie(NULL), cout.tie(NULL);
  ;
  cin >> m;
  for (int i = 0; i < m; ++i) {
    cin >> a[i];
    if (a[i] % 2 == 0) {
      even.push_back(make_pair(a[i], i));
    } else {
      odd.push_back(make_pair(a[i], i));
    }
  }
  if (((int)(even).size()) != ((int)(odd).size())) {
    cout << "Impossible";
    return 0;
  }
  m /= 2;
  n = 2 * m + 2;
  s = 0;
  t = 2 * m + 1;
  for (int i = 1; i <= m; ++i) {
    add_edge(s, i, 2);
  }
  for (int i = m + 1; i <= 2 * m; ++i) {
    add_edge(i, t, 2);
  }
  for (int i = 0; i < m; ++i) {
    for (int j = 0; j < m; ++j) {
      if (prime(even[i].first + odd[j].first)) {
        add_edge(i + 1, j + m + 1, 1);
      }
    }
  }
  dinic();
  int size = 0;
  for (int i = 0; i < ((int)(e).size()); ++i) {
    if (e[i].a == s && e[i].b >= 1 && e[i].b <= m) {
      size += e[i].flow;
    }
  }
  if (size != 2 * m) {
    cout << "Impossible";
    return 0;
  }
  for (int i = 0; i < ((int)(e).size()); ++i) {
    if (IF(e[i].a, 1, m) && IF(e[i].b, m + 1, 2 * m) && e[i].flow == 1) {
      c[e[i].a].push_back(e[i].b);
      c[e[i].b].push_back(e[i].a);
    }
  }
  for (int i = 1; i <= m; ++i) {
    if (!used[i]) {
      dfs(i);
      cnt++;
    }
  }
  cout << cnt << '\n';
  for (int i = 0; i < cnt; ++i) {
    cout << ((int)(ans[i]).size());
    for (int j = 0; j < ((int)(ans[i]).size()); ++j) {
      cout << " " << ans[i][j];
    }
    cout << '\n';
  }
  return 0;
}
int32_t main() {
  solve();
  return 0;
}
