Problem: Consider an undirected graph G with n vertices. There is a value a_i in each vertex.

Two vertices i and j are connected with an edge if and only if gcd(a_i, a_j) > 1, where gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y.

Consider a set of vertices. Let's call a vertex in this set fair if it is connected with an edge with all other vertices in this set.

You need to find a set of k vertices (where k is a given integer, 2 ⋅ k ≤ n) where all vertices are fair or all vertices are not fair. One can show that such a set always exists.

Input

The first line contains integers n and k (6 ≤ 2 ⋅ k ≤ n ≤ 10^5) — the number of vertices and parameter k.

The second line contains n integers a_1, a_2, …, a_n (2 ≤ a_i ≤ 10^7) — the values in the vertices.

Output

Print exactly k distinct integers — the indices of the vertices in the chosen set in any order.

Examples

Input


6 3
6 15 10 8 14 12


Output


2 4 5

Input


8 4
11 15 10 6 21 15 10 6


Output


5 7 1 2

Input


10 5
3003 17017 3230 49742 546 41990 17765 570 21945 36465


Output


1 2 4 5 6

Note

In the first test case, set \{2, 4, 5\} is an example of set where no vertices are fair. The vertex 2 does not share an edge with vertex 4 since gcd(15, 8) = 1. The vertex 4 does not share an edge with vertex 2. The vertex 5 does not share an edge with vertex 2.

In the second test case, set \{8, 5, 6, 4\} is an example of a set where all vertices are fair.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5, V = 1e7 + 1;
int n, K, m;
int a[N];
int pri[V], mp[V], mu[V];
void Euler(int n) {
  static bitset<V> vis;
  int num = 0;
  mu[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (!vis.test(i)) {
      pri[num++] = i;
      mp[i] = i;
      mu[i] = -1;
    }
    for (int *p = pri, *_ = pri + num, x; p != _ && (x = *p * i) <= n; p++) {
      vis.set(x);
      mp[x] = *p;
      if (i % *p) {
        mu[x] = -mu[i];
      } else {
        mu[x] = 0;
        break;
      }
    }
  }
}
vector<int> factor(int x) {
  vector<int> P;
  while (x > 1) {
    P.push_back(mp[x]);
    while (!(x % P.back())) x /= P.back();
  }
  return P;
}
vector<int> Ps(vector<int> P) {
  vector<int> rt;
  int n = P.size();
  for (int s = 0; s < (1 << n); s++) {
    int prod = 1;
    for (int j = 0; j < s && (s >> j); j++) {
      if ((s >> j) & 1) {
        prod *= P[j];
      }
    }
    rt.push_back(prod);
  }
  return rt;
}
int cnt[V];
vector<int> P[N];
void add(vector<int>& P) {
  for (auto idx : P) cnt[idx] += mu[idx];
}
void clear(vector<int>& P) {
  for (auto idx : P) cnt[idx] = 0;
}
int query(vector<int>& P) {
  int rt = 0;
  for (auto idx : P) rt += cnt[idx];
  return rt;
}
vector<int> cli, ncli;
void add(int l, int r) {
  for (int i = l; i <= r; i++) {
    add(P[cli[i]]);
  }
}
void clear(int l, int r) {
  for (int i = l; i <= r; i++) {
    clear(P[cli[i]]);
  }
}
int cost[N];
bool in_cli[N];
vector<int> adj[N];
vector<int> L1, L2;
void dividing(int l, int r, vector<int> S) {
  if (l == r) {
    adj[cli[l]] = S;
    for (auto x : S) adj[x].push_back(cli[l]);
    return;
  }
  int mid = (l + r) >> 1;
  add(l, mid);
  vector<int> vL, vR;
  for (auto x : S) {
    if (query(P[x])) {
      vL.push_back(x);
    } else {
      vR.push_back(x);
    }
  }
  clear(l, mid);
  (vector<int>()).swap(S);
  dividing(l, mid, vL);
  dividing(mid + 1, r, vR);
}
int sx, sy;
bool get(int l, int r, int x) {
  add(l, r);
  if (!query(P[x])) {
    clear(l, r);
    return false;
  }
  clear(l, r);
  int mid;
  while (l < r) {
    mid = (l + r) >> 1;
    for (int i = l; i <= mid; i++) {
      add(P[cli[i]]);
    }
    bool flag = query(P[x]);
    for (int i = l; i <= mid; i++) {
      clear(P[cli[i]]);
    }
    if (flag) {
      r = mid;
    } else {
      l = mid + 1;
    }
  }
  sx = cli[l];
  sy = x;
  return true;
}
int main() {
  scanf("%d%d", &n, &K);
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
    m = max(m, a[i]);
  }
  Euler(m);
  for (int i = 1; i <= n && (signed)cli.size() < K; i++) {
    P[i] = Ps(factor(a[i]));
    if ((in_cli[i] = !query(P[i]))) {
      add(P[i]);
      cli.push_back(i);
    } else {
      ncli.push_back(i);
    }
  }
  if ((signed)cli.size() == K) {
    for (auto x : cli) {
      printf("%d ", x);
    }
    return 0;
  }
  for (auto x : cli) {
    clear(P[x]);
  }
  dividing(0, (signed)cli.size() - 1, ncli);
  for (auto x : ncli) {
    cost[x] = 2;
    L2.push_back(x);
  }
  vector<int> ans;
  if ((K & 1)) {
    bool suc = false;
    int s = (signed)cli.size();
    add(0, s - 1);
    for (auto x : ncli) {
      if (query(P[x]) > 1) {
        clear(0, s - 1);
        int y = adj[x][0], idx = 0;
        while (cli[idx] ^ y) idx++;
        if (idx) {
          suc = get(0, idx - 1, x);
        }
        if (idx < s - 1 && !suc) {
          suc = get(idx + 1, s - 1, x);
        }
        break;
      }
    }
    if (suc) {
      cost[sy] = -1;
      K -= 3;
      ans.push_back(sx);
      ans.push_back(sy);
      ans.push_back(adj[sy][0]);
      for (auto z : adj[sx]) {
        if (cost[z] == 2) {
          cost[z] = 1;
          L1.push_back(z);
        }
      }
      for (auto z : adj[adj[sy][0]]) {
        if (cost[z] == 2) {
          cost[z] = 1;
          L1.push_back(z);
        }
      }
    } else {
      clear(0, s - 1);
    }
  }
  while (!L2.empty() && K >= 2) {
    int x = L2.back();
    L2.pop_back();
    if (cost[x] < 2) continue;
    K -= 2;
    int y = adj[x][0];
    ans.push_back(x);
    ans.push_back(y);
    for (auto z : adj[y]) {
      cost[z] = 1;
      if (z != x) {
        L1.push_back(z);
      }
    }
  }
  while (!L1.empty() && K) {
    ans.push_back(L1.back());
    L1.pop_back();
    K--;
  }
  assert(!K);
  for (auto x : ans) {
    printf("%d ", x);
  }
  return 0;
}