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 JAVA):

Solution:
import java.io.*;
import java.util.*;

public class CF1249D2 extends PrintWriter {
	CF1249D2() { super(System.out, true); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF1249D2 o = new CF1249D2(); o.main(); o.flush();
	}

	int[][] xx;
	int[] pq, iq; int cnt;
	static class V {
		int i, lr;
		V(int i, int lr) {
			this.i = i; this.lr = lr;
		}
	}
	boolean less(int u, int v) { return xx[u][1] > xx[v][1]; }
	int i2(int i) {
		return (i *= 2) > cnt ? 0 : i < cnt && less(pq[i + 1], pq[i]) ? i + 1 : i;
	}
	void pq_up(int u) {
		int i, j, v;
		for (i = iq[u]; (j = i / 2) > 0 && less(u, v = pq[j]); i = j)
			pq[iq[v] = i] = v;
		pq[iq[u] = i] = u;
	}
	void pq_dn(int u) {
		int i, j, v;
		for (i = iq[u]; (j = i2(i)) > 0 && less(v = pq[j], u); i = j)
			pq[iq[v] = i] = v;
		pq[iq[u] = i] = u;
	}
	void pq_add(int u) {
		iq[u] = ++cnt; pq_up(u);
	}
	int pq_remove_first() {
		int u = pq[1], v = pq[cnt--];
		if (v != u) {
			iq[v] = 1; pq_dn(v);
		}
		return u;
	}
	void pq_remove(int u) {
		int v = pq[cnt--];
		if (v != u) {
			iq[v] = iq[u]; pq_dn(v); pq_up(v);
		}
	}
	void main() {
		int n = sc.nextInt();
		int k = sc.nextInt();
		xx = new int[n][2];
		V[] vv = new V[n * 2];
		for (int i = 0; i < n; i++) {
			xx[i][0] = sc.nextInt();
			xx[i][1] = sc.nextInt();
			vv[i << 1] = new V(i, 0);
			vv[i << 1 | 1] = new V(i, 1);
		}
		Arrays.sort(vv, (u, v) -> {
			int xu = xx[u.i][u.lr];
			int xv = xx[v.i][v.lr];
			return xu != xv ? xu - xv : u.lr - v.lr;
		});
		boolean[] removed = new boolean[n];
		pq = new int[1 + n]; iq = new int[n];
		for (int i = 0, j; i < n * 2; i = j) {
			V u = vv[i];
			int x = xx[u.i][u.lr];
			j = i;
			while (j < n * 2) {
				V v = vv[j];
				if (xx[v.i][v.lr] == x && v.lr == 0) {
					pq_add(v.i);
					j++;
				} else
					break;
			}
			while (cnt > k) {
				int h = pq_remove_first();
				removed[h] = true;
			}
			while (j < n * 2) {
				V v = vv[j];
				if (xx[v.i][v.lr] == x && v.lr == 1) {
					if (!removed[v.i])
						pq_remove(v.i);
					j++;
				} else
					break;
			}
		}
		int m = 0;
		for (int i = 0; i < n; i++)
			if (removed[i])
				m++;
		println(m);
		for (int i = 0; i < n; i++)
			if (removed[i])
				print(i + 1 + " ");
		println();
	}
}