CodeforcesAug 13, 2025

Points in Segments

Hazrat Ali

Codeforces

You are given a set of n segments on the axis Ox, each segment has integer endpoints between 1 and m inclusive. Segments may intersect, overlap or even coincide with each other. Each segment is characterized by two integers li and ri (1lirim) — coordinates of the left and of the right endpoints.

Consider all integer points between 1 and m inclusive. Your task is to print all such points that don't belong to any segment. The point x belongs to the segment [l;r] if and only if lxr.

Input

The first line of the input contains two integers n and m (1n,m100) — the number of segments and the upper bound for coordinates.

The next n lines contain two integers each li and ri (1lirim) — the endpoints of the i-th segment. Segments may intersect, overlap or even coincide with each other. Note, it is possible that li=ri, i.e. a segment can degenerate to a point.

Output

In the first line print one integer k — the number of points that don't belong to any segment.

In the second line print exactly k integers in any order — the points that don't belong to any segment. All points you print should be distinct.

If there are no such points at all, print a single integer 0 in the first line and either leave the second line empty or do not print it at all.

Examples
Input
3 5
2 2
1 2
5 5
Output
2
3 4
Input
1 7
1 7
Output
0

Solution

#include <bits/stdc++.h>
using namespace std;

int main() {
  int n, m;
  cin >> n >> m;
  vector<int> s(m);
  for (int i = 0; i < n; i++) {
    int l, r;
    cin >> l >> r;
    for (int j = l - 1; j < r; j++) {
      s[j]++;
    }
  }
  int cnt = 0;
  for (int i = 0; i < m; i++) {
    if (s[i] == 0) {
      cnt++;
    }
  }
  cout << cnt << "\n";
  for (int i = 0; i < m; i++) {
    if (s[i] == 0) {
      cout << i + 1 << " ";
    }
  }
  cout << "\n";
  return 0;
}








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