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;
const int dr[] = {-1, 1, 0, 0, -1, -1, 1, 1};
const int dc[] = {0, 0, 1, -1, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const int INF = 1e9;
const int N = 1e5 + 5;
const long double pi = acos(-1);
const double eps = 1e-9;
int n, m, st, ed, cnt;
pair<int, int> a[N];
vector<int> ans;
vector<pair<int, int> > lst;
bool vis[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) cin >> a[i].first >> a[i].second;
  sort(a + 1, a + 1 + n);
  for (long long i = 1; i <= n; i++) {
    if (i == 1) {
      st = a[i].first;
      ed = a[i].second;
    } else {
      if (a[i].first <= ed) {
        ed = max(ed, a[i].second);
      } else {
        lst.push_back({st, ed});
        st = a[i].first;
        ed = a[i].second;
      }
    }
  }
  lst.push_back({st, ed});
  int sz = lst.size();
  for (long long i = 0; i < sz; i++) {
    for (long long j = lst[i].first; j <= lst[i].second; j++) {
      vis[j] = 1;
    }
  }
  for (long long i = 1; i <= m; i++) {
    if (!vis[i]) cnt++;
  }
  cout << cnt << '\n';
  for (long long i = 1; i <= m; i++) {
    if (!vis[i]) cout << i << ' ';
  }
  cout << '\n';
  return 0;
}