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>
const int MAXN = 1e5 + 7;
const int MAXM = 2e5 + 7;
const int inf = 1e9 + 7;
using namespace std;
queue<int> que;
int n, a[MAXN], tot, maxflow;
int pri[MAXN], isp[MAXN];
int ccnt, home[MAXN], nxt[MAXM], to[MAXM], w[MAXM], h[MAXN], cur[MAXN];
int vis[MAXN], T, s, t;
int flg[MAXN];
vector<int> ans[500];
inline void cuntu(int x, int y, int z) {
  ccnt++;
  nxt[ccnt] = home[x];
  home[x] = ccnt;
  to[ccnt] = y;
  w[ccnt] = z;
}
inline void ct(int x, int y, int z) {
  cuntu(x, y, z);
  cuntu(y, x, 0);
}
inline void init() {
  for (int i = 2; i < MAXN; ++i) {
    if (!isp[i]) {
      pri[++tot] = i;
    }
    for (int j = 1; j <= tot && i * pri[j] < MAXN; ++j) {
      isp[i * pri[j]] = 1;
      if (i % pri[j] == 0) {
        break;
      }
    }
  }
  return;
}
inline void build() {
  s = 2 * n + 1, t = 2 * n + 2;
  ccnt = 1;
  for (int i = 1; i <= n; ++i) {
    if (a[i] & 1)
      ct(s, i, 2);
    else
      ct(i + n, t, 2);
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      if (!isp[a[i] + a[j]] && (a[i] & 1)) {
        ct(i, j + n, 1);
      }
    }
  }
  return;
}
inline int bfs() {
  memset(h, 0, sizeof(h));
  while (!que.empty()) que.pop();
  que.push(s);
  h[s] = 1;
  while (!que.empty()) {
    int u = que.front();
    que.pop();
    for (int i = home[u]; i; i = nxt[i]) {
      int v = to[i];
      if (!h[v] && w[i] > 0) {
        h[v] = h[u] + 1;
        que.push(v);
      }
    }
    if (h[t]) break;
  }
  return h[t];
}
inline int dfs(int u, int nflow) {
  if (nflow == 0 || u == t) return nflow;
  int ret = nflow, a;
  for (int &i = cur[u]; i; i = nxt[i]) {
    int v = to[i];
    if (h[v] == h[u] + 1 && (a = dfs(v, min(ret, w[i])))) {
      ret -= a;
      w[i] -= a;
      w[i ^ 1] += a;
      if (!ret) break;
    }
  }
  if (ret == nflow) h[u] = -1;
  return (nflow - ret);
}
inline void Dinic() {
  maxflow = 0;
  while (bfs()) {
    memcpy(cur, home, sizeof(cur));
    maxflow += dfs(s, inf);
  }
  return;
}
int st[MAXN], tp;
inline void dfs1(int u, int F) {
  bool tmp;
  if (u > n)
    tmp = 1;
  else
    tmp = 0;
  flg[u] = 1;
  st[++tp] = u > n ? u - n : u;
  for (int i = home[u]; i; i = nxt[i]) {
    int v = to[i];
    if (tmp ^ w[i]) continue;
    if (v == F || v == s || v == t) continue;
    if (flg[v]) return;
    dfs1(v, u);
  }
  return;
}
inline void solve() {
  if (maxflow != n) {
    return (void)puts("Impossible");
  }
  tot = 0;
  for (int i = 1; i <= n; ++i) {
    if (!flg[i] && (a[i] & 1)) {
      dfs1(i, 0);
      ++tot;
      while (tp) {
        ans[tot].push_back(st[tp]);
        tp--;
      }
    }
  }
  printf("%d\n", tot);
  for (int i = 1; i <= tot; ++i) {
    printf("%d ", (int)ans[i].size());
    for (int j = 0; j < (int)ans[i].size(); ++j) {
      printf("%d ", ans[i][j]);
    }
    puts("");
  }
  return;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
  }
  init();
  build();
  Dinic();
  solve();
  return 0;
}