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.
#include <bits/stdc++.h>
using namespace std;
bool visited[110];
void setVisited(int L, int R) {
  for (int i = L; i <= R; i++) visited[i] = true;
  return;
}
vector<int> getAnswer(int M) {
  vector<int> ans;
  for (int i = 1; i <= M; i++)
    if (!visited[i]) ans.push_back(i);
  return ans;
}
void showAnswer(vector<int> output) {
  cout << output.size() << endl;
  for (int i = 0; i < output.size(); i++) cout << output[i] << " ";
  cout << endl;
}
int main() {
  long long int N, M;
  cin >> N >> M;
  for (int i = 0; i < N; i++) {
    long long int L, R;
    cin >> L >> R;
    setVisited(L, R);
  }
  showAnswer(getAnswer(M));
}
