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;
  cin >> n;
  int m;
  cin >> m;
  int a, b;
  vector<pair<int, int> > l;
  for (int i = 0; i < n; i++) {
    cin >> a >> b;
    l.push_back(make_pair(a, b));
  }
  sort(l.begin(), l.end());
  int x = 0, i, maxr = l[x].second;
  vector<int> ans;
  for (i = 1; i < m; i++) {
    if (i < l[x].first)
      ans.push_back(i);
    else
      break;
  }
  for (x = 1; x < n; x++) {
    if (l[x].first > l[x - 1].second) {
      int z = l[x - 1].second + 1;
      while (z < l[x].first) {
        if (z > maxr) ans.push_back(z);
        z++;
      }
    }
    if (l[x].second > maxr) maxr = l[x].second;
  }
  maxr++;
  while (maxr <= m) {
    ans.push_back(maxr);
    maxr++;
  }
  cout << ans.size() << endl;
  for (i = 0; i < ans.size(); i++) cout << ans[i] << " ";
}