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;
const int maxn = 2e5 + 5;
const int INF = 0x3f3f3f3f;
const long long mod = 1e9 + 7;
int lazy[maxn << 2];
int n, k, way[maxn];
struct node {
  int l, r;
  int num;
} a[maxn];
priority_queue<node> q;
bool operator<(node a, node b) { return a.r < b.r; }
bool cmp(node a, node b) {
  if (a.l == b.l) return a.r > b.r;
  return a.l < b.l;
}
struct Tree {
  int l, r;
  int sum;
  int mid() { return (l + r) >> 1; }
} tree[maxn << 2];
void PushUp(int rt) {
  tree[rt].sum = tree[rt << 1].sum + tree[rt << 1 | 1].sum;
}
void PushDown(int rt, int m) {
  if (lazy[rt]) {
    lazy[rt << 1] += lazy[rt];
    lazy[rt << 1 | 1] += lazy[rt];
    tree[rt << 1].sum += lazy[rt] * (m - (m >> 1));
    tree[rt << 1 | 1].sum += lazy[rt] * (m >> 1);
    lazy[rt] = 0;
  }
}
void build(int l, int r, int rt) {
  tree[rt].l = l;
  tree[rt].r = r;
  lazy[rt] = 0;
  if (l == r) {
    tree[rt].sum = 0;
    return;
  }
  int m = tree[rt].mid();
  build(l, m, (rt << 1));
  build(m + 1, r, (rt << 1 | 1));
  PushUp(rt);
}
void update(int c, int l, int r, int rt) {
  if (tree[rt].l == l && tree[rt].r == r) {
    lazy[rt] += c;
    tree[rt].sum += c * (r - l + 1);
    return;
  }
  if (tree[rt].l == tree[rt].r) return;
  int m = tree[rt].mid();
  PushDown(rt, tree[rt].r - tree[rt].l + 1);
  if (r <= m)
    update(c, l, r, rt << 1);
  else if (l > m)
    update(c, l, r, rt << 1 | 1);
  else {
    update(c, l, m, rt << 1);
    update(c, m + 1, r, rt << 1 | 1);
  }
  PushUp(rt);
}
int Query(int l, int r, int rt) {
  if (l == tree[rt].l && r == tree[rt].r) {
    return tree[rt].sum;
  }
  int m = tree[rt].mid();
  PushDown(rt, tree[rt].r - tree[rt].l + 1);
  int res = 0;
  if (r <= m)
    res = max(res, Query(l, r, rt << 1));
  else if (l > m)
    res = max(res, Query(l, r, rt << 1 | 1));
  else {
    res = max(res, Query(l, m, rt << 1));
    res = max(res, Query(m + 1, r, rt << 1 | 1));
  }
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  build(1, 200000, 1);
  for (int i = 0; i < n; i++) {
    cin >> a[i].l >> a[i].r;
    a[i].num = i + 1;
    update(1, a[i].l, a[i].r, 1);
  }
  sort(a, a + n, cmp);
  int ans = 0, pos = 0;
  for (int i = 1; i <= 200000; i++) {
    if (pos >= n) break;
    while (!q.empty() && q.top().r < i) q.pop();
    while (a[pos].r >= i && a[pos].l <= i) {
      q.push({a[pos].l, a[pos].r, a[pos].num});
      pos++;
    }
    while (Query(i, i, 1) > k) {
      node now = q.top();
      q.pop();
      update(-1, now.l, now.r, 1);
      way[ans++] = now.num;
    }
  }
  cout << ans << '\n';
  for (int i = 0; i < ans; i++) cout << way[i] << " ";
  return 0;
}