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;
vector<long long> graph[100000];
long long vis[100000], c[100000], dis[100000];
void c_p_c() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
long long dfs(long long i, long long co, long long m) {
  vis[i] = 1;
  long long ans = 0;
  if (c[i] == 1)
    co++;
  else
    co = 0;
  if (co > m) return 0;
  if (i != 1 && graph[i].size() == 1) return 1;
  for (long long j = 0; j < graph[i].size(); j++) {
    if (vis[graph[i][j]] == 0) ans += dfs(graph[i][j], co, m);
  }
  return ans;
}
void dfsofgraph(long long i, long long di) {
  vis[i] = 1;
  dis[i] = di;
  for (auto &it : graph[i]) {
    if (!(vis[it])) dfsofgraph(it, dis[i] + 1);
  }
}
int32_t main() {
  c_p_c();
  long long n, m;
  cin >> n >> m;
  set<long long> se;
  long long i;
  while (n--) {
    long long x, y;
    cin >> x >> y;
    for (i = x; i <= y; i++) se.insert(i);
  }
  set<long long> res;
  cout << m - se.size() << endl;
  auto it = se.begin();
  for (i = 1; i <= m; i++) {
    if (se.find(i) != se.end())
      continue;
    else
      cout << i << " ";
  }
}