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;
long long mod = 1e9 + 7;
long long inf = 1e17;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, m;
  cin >> n >> m;
  long long l[n], r[n];
  for (long long i = 0; i < n; i++) {
    cin >> l[i] >> r[i];
  }
  long long d[m + 1];
  memset(d, 0, sizeof(d));
  long long ans = 0;
  for (long long i = 0; i < n; i++) {
    for (long long j = l[i]; j <= r[i]; j++) {
      if (d[j] == 0) ans++;
      d[j] = 1;
    }
  }
  ans = m - ans;
  cout << ans << "\n";
  for (long long i = 1; i <= m; i++) {
    if (d[i] == 0) cout << i << " ";
  }
  cout << "\n";
}
