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;
using ll = long long int;
using P = pair<int, int>;
using T = tuple<int, int, int>;
using edge = struct { int to, cost; };
const int MOD = 1e9 + 7;
const int iINF = 1e9;
const long long int llINF = 1e18;
const double PI = acos(-1.0);
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int N, M;
  cin >> N >> M;
  vector<bool> vec(M, false);
  for (int i = 0; i < (N); i++) {
    int l, r;
    cin >> l >> r;
    l--;
    for (int j = l; j < (r); j++) {
      vec[j] = true;
    }
  }
  vector<int> ans;
  for (int i = 0; i < (M); i++) {
    if (vec[i]) continue;
    ans.push_back(i + 1);
  }
  cout << (int)(ans).size() << endl;
  for (auto x : ans) {
    cout << x << " ";
  }
  cout << endl;
  return 0;
}
