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 arr[1005], arr2[1005];
int n, m;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0), cout.precision(15);
  while (cin >> n >> m) {
    vector<int> v;
    for (int i = int(0); i < int(n); i++) cin >> arr[i] >> arr2[i];
    for (int i = int(1); i < int(m + 1); i++) {
      bool b = 1;
      for (int j = int(0); j < int(n); j++) {
        if (arr[j] <= i && i <= arr2[j]) {
          b = 0;
        }
      }
      if (b) v.push_back(i);
    }
    cout << v.size() << endl;
    if (v.size()) cout << v[0];
    for (int i = int(1); i < int(v.size()); i++) {
      cout << " " << v[i];
    }
    if (v.size()) {
      cout << endl;
    }
  }
  return 0;
}