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() {
  long long int n, m, i, j = 2, k = 0, t, a = 5, b, c, ck = 0, dk = 0;
  scanf("%lld %lld", &n, &m);
  long long int arr[n], brr[n], crr[m + 1], drr[m + 1];
  memset(crr, 0, sizeof(crr));
  for (i = 1; i <= n; i++) {
    scanf("%lld %lld", &arr[i], &brr[i]);
    for (j = arr[i]; j <= brr[i]; j++) {
      crr[j] = 1;
    }
  }
  for (i = 1; i <= m; i++) {
    if (crr[i] != 1) {
      ck++;
    }
  }
  cout << ck << endl;
  for (i = 1; i <= m; i++)
    if (crr[i] != 1) cout << i << " ";
}