Problem: The only difference between easy and hard versions is constraints.

You are given n segments on the coordinate axis OX. Segments can intersect, lie inside each other and even coincide. The i-th segment is [l_i; r_i] (l_i ≤ r_i) and it covers all integer points j such that l_i ≤ j ≤ r_i.

The integer point is called bad if it is covered by strictly more than k segments.

Your task is to remove the minimum number of segments so that there are no bad points at all.

Input

The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 2 ⋅ 10^5) — the number of segments and the maximum number of segments by which each integer point can be covered.

The next n lines contain segments. The i-th line contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ 2 ⋅ 10^5) — the endpoints of the i-th segment.

Output

In the first line print one integer m (0 ≤ m ≤ n) — the minimum number of segments you need to remove so that there are no bad points.

In the second line print m distinct integers p_1, p_2, ..., p_m (1 ≤ p_i ≤ n) — indices of segments you remove in any order. If there are multiple answers, you can print any of them.

Examples

Input


7 2
11 11
9 11
7 8
8 9
7 8
9 11
7 9


Output


3
4 6 7 


Input


5 1
29 30
30 30
29 29
28 30
30 30


Output


3
1 4 5 


Input


6 1
2 3
3 3
2 3
2 2
2 3
2 3


Output


4
1 3 5 6 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second < b.second);
}
const long long INF = 1e10 + 1;
const int mod = 1e9 + 7;
inline int add(int x, int y) {
  x += y;
  if (x >= mod) return x - mod;
  return x;
}
inline int sub(int x, int y) {
  x -= y;
  if (x < 0) return x + mod;
  return x;
}
inline int mul(int x, int y) { return (x * 1ll * y) % mod; }
inline int power(int x, int y) {
  int ans = 1;
  while (y) {
    if (y & 1) ans = mul(ans, x);
    x = mul(x, x);
    y >>= 1;
  }
  return ans;
}
inline int inv(int x) { return power(x, mod - 2); }
const long long FACTORIAL_SIZE = 1.1e6;
long long fact[FACTORIAL_SIZE], ifact[FACTORIAL_SIZE];
void gen_factorial(long long n) {
  fact[0] = fact[1] = ifact[0] = ifact[1] = 1;
  for (long long i = 2; i <= n; i++) {
    fact[i] = (i * fact[i - 1]) % mod;
  }
  ifact[n] = inv(fact[n]);
  for (long long i = n - 1; i >= 2; i--) {
    ifact[i] = ((i + 1) * ifact[i + 1]) % mod;
  }
}
long long nck(long long n, long long k) {
  long long den = (ifact[k] * ifact[n - k]) % mod;
  return (den * fact[n]) % mod;
}
const int N = 2e5 + 5;
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<pair<long long, long long>> v(n);
  vector<long long> cnt(N);
  vector<vector<long long>> ev(N);
  for (long long i = 0; i < n; i++) {
    cin >> v[i].first >> v[i].second;
    cnt[v[i].first]++;
    cnt[v[i].second + 1]--;
    ev[v[i].first].push_back(i + 1);
    ev[v[i].second + 1].push_back(-i - 1);
  }
  for (long long i = 1; i < N; i++) {
    cnt[i] += cnt[i - 1];
  }
  vector<long long> ans(N), sub(N);
  set<pair<long long, long long>> curSegs;
  long long curSub = 0;
  for (long long i = 0; i < N; i++) {
    curSub += sub[i];
    for (auto it : ev[i]) {
      if (it > 0) {
        curSegs.insert({v[it - 1].second, it - 1});
      } else {
        auto it2 = curSegs.find({v[-it - 1].second, -it - 1});
        if (it2 != curSegs.end()) curSegs.erase(it2);
      }
    }
    while (cnt[i] - curSub > k) {
      assert(!curSegs.empty());
      long long pos = curSegs.rbegin()->second;
      curSegs.erase(prev(curSegs.end()));
      curSub++;
      sub[v[pos].second + 1]--;
      ans[pos] = 1;
    }
  }
  cout << accumulate(ans.begin(), ans.end(), 0) << "\n";
  for (long long i = 0; i < n; i++) {
    if (ans[i]) cout << i + 1 << " ";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long tc = 1;
  long long i = 1;
  while (tc--) {
    solve();
    i++;
  }
  return 0;
}