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;
long long n, m, sq[100069], zs = 0;
pair<long long, long long> a[100069];
int main() {
  long long i, j, mx = -1e18;
  scanf("%lld%lld", &n, &m);
  for (i = 0; i < n; i++) {
    scanf("%lld%lld", &a[i].first, &a[i].second);
  }
  sort(a, a + n);
  for (j = 0, i = 1; i <= m; i++) {
    for (; j < n && a[j].first <= i; j++) {
      mx = max(mx, a[j].second);
    }
    if (i > mx) {
      zs++;
      sq[zs] = i;
    }
  }
  printf("%lld\n", zs);
  for (i = 1; i <= zs; i++) {
    printf("%lld%c", sq[i], " \n"[i == zs]);
  }
}