Problem: 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 l_i and r_i (1 ≤ l_i ≤ r_i ≤ m) — 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 l ≤ x ≤ r.

Input

The first line of the input contains two integers n and m (1 ≤ n, m ≤ 100) — the number of segments and the upper bound for coordinates.

The next n lines contain two integers each l_i and r_i (1 ≤ l_i ≤ r_i ≤ m) — the endpoints of the i-th segment. Segments may intersect, overlap or even coincide with each other. Note, it is possible that l_i=r_i, 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

Note

In the first example the point 1 belongs to the second segment, the point 2 belongs to the first and the second segments and the point 5 belongs to the third segment. The points 3 and 4 do not belong to any segment.

In the second example all the points from 1 to 7 belong to the first segment.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, m;
  cin >> n >> m;
  vector<pair<int, int>> v;
  for (int i = 0; i < n; i++) {
    int l, r;
    cin >> l >> r;
    pair<int, int> p;
    p.first = l;
    p.second = r;
    v.push_back(p);
  }
  sort(v.begin(), v.end());
  vector<int> ans;
  int d = v[0].first - 1;
  int maximum = v[0].second;
  for (int i = 0; i < d; i++) {
    ans.push_back(i + 1);
  }
  for (int i = 1; i < n; i++) {
    if (v[i].first >= maximum) {
      int diff = v[i].first - maximum;
      for (int i = 0; i < diff - 1; i++) {
        ans.push_back(maximum + i + 1);
      }
      maximum = v[i].second;
    } else {
      if (v[i].second > maximum) {
        maximum = v[i].second;
      }
    }
  }
  if (maximum < m) {
    int diff = m - maximum;
    for (int i = maximum + 1; i <= m; i++) {
      ans.push_back(i);
    }
  }
  int count = ans.size();
  cout << count << endl;
  for (int i = 0; i < count; i++) {
    cout << ans[i] << " ";
  }
}