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;
const int N = (int)2e5 + 5;
const int INF = (int)1e9 + 7;
int dx[8] = {1, 0, -1, 0, -1, -1, 1, 1};
int dy[8] = {0, 1, 0, -1, -1, 1, -1, 1};
int ini() {
  int x;
  scanf("%d", &x);
  return x;
}
long long inl() {
  long long x;
  scanf("%lld", &x);
  return x;
}
int a[100009];
bool vis[1009];
int main() {
  int n, i, j, m;
  n = ini();
  m = ini();
  for (i = 1; i <= n; i++) {
    int x, y;
    cin >> x >> y;
    for (int j = x; j <= y; j++) {
      vis[j] = true;
    }
  }
  int res = 0;
  j = 1;
  for (int i = 1; i <= m; i++) {
    if (!vis[i]) {
      res++;
      a[j++] = i;
    }
  }
  cout << res << endl;
  for (int i = 1; i < j; i++) cout << a[i] << " ";
  return 0;
}
