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;
void File_input() {
  freopen("input.txt", "r", stdin);
  freopen("output.txt", "w", stdout);
}
void FastInputOutput() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
inline long long readLong(bool t = 0) {
  long long tmp;
  if (!t) {
    cin >> tmp;
    return tmp;
  }
  scanf("%lld", &tmp);
  return tmp;
}
inline int readInt(bool t = 0) {
  int tmp;
  if (!t) {
    cin >> tmp;
    return tmp;
  }
  scanf("%d", &tmp);
  return tmp;
}
int Abs(int n) {
  if (n < 0) return -1 * n;
  return n;
}
long long Abs(long long n) {
  if (n < 0) return -1 * n;
  return n;
}
int main() {
  vector<pair<int, int> > ans;
  vector<int> a;
  int n, m, l, r;
  cin >> n >> m;
  while (n--) {
    cin >> l >> r;
    ans.push_back({l, r});
  }
  for (int j = 1; j <= m; j++) {
    bool f = 0;
    for (int i = 0; i < ans.size(); i++) {
      if (ans[i].first <= j && j <= ans[i].second) {
        f = 0;
        break;
      } else {
        f = 1;
      }
    }
    if (f) a.push_back(j);
  }
  cout << a.size() << "\n";
  for (auto i : a) cout << i << " ";
  cerr << "\nTime elapsed " << 1.0 * clock() / CLOCKS_PER_SEC << " secs";
  return 0;
}