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 longestIncreasingSubsequenceDP(int a[], int n) {
  int dp[n], ans = 0;
  for (int k = 0; k < n; k++) {
    dp[k] = 1;
    for (int i = 0; i < k; i++) {
      if (a[i] <= a[k]) {
        dp[k] = max(dp[k], dp[i] + 1);
        ans = max(dp[k], ans);
      }
    }
  }
  return ans;
}
int gcd(int a, int b) {
  if (a == 0) return b;
  if (b == 0) return a;
  if (a == b) return a;
  if (a > b) return gcd(a - b, b);
  return gcd(a, b - a);
}
void scanArray(int arr[], int n) {
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
}
int highestFromThree(int n1, int n2, int n3) { return max(max(n1, n2), n3); }
int highestFromFour(int n1, int n2, int n3, int n4) {
  return max(max(n1, n2), max(n3, n4));
}
int result[1000] = {0};
int factDP(int n) {
  if (n >= 0) {
    result[0] = 1;
    for (int i = 1; i <= n; ++i) {
      result[i] = i * result[i - 1];
    }
    return result[n];
  }
}
int main() {
  int n, m;
  cin >> n >> m;
  int l[n], r[n];
  vector<pair<int, int>> v;
  int points[m];
  for (int i = 0; i < m; i++) {
    points[i] = i + 1;
  }
  for (int i = 0; i < n; i++) {
    cin >> l[i] >> r[i];
    v.push_back(make_pair(l[i], r[i]));
  }
  int count = m;
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < v.size(); j++) {
      if (points[i] >= v[j].first && points[i] <= v[j].second) {
        count--;
        points[i] = 0;
        break;
      }
    }
  }
  cout << count << "\n";
  for (int i = 0; i < m; i++) {
    if (points[i] != 0) {
      cout << points[i] << " ";
    }
  }
  return 0;
}