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;
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
const int MAXN = 205;
int n, m;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a * (b / gcd(a, b)); }
long long modexp(long long a, long long b, long long M) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % M;
    b >>= 1;
    a = (a * a) % M;
  }
  return res;
}
bool is_prime(int num) {
  if (num % 2 == 0 && num > 2) {
    return 0;
  }
  for (int i = 3; i * i <= num; i += 2) {
    if (num % i == 0) {
      return 0;
    }
  }
  return 1;
}
int l, r;
vector<int> res;
vector<pair<int, int> > seg;
bool vis[1000];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    cin >> l >> r;
    for (int i = l; i < r + 1; i++) vis[i] = true;
  }
  for (int i = 1; i < m + 1; i++) {
    if (!vis[i]) {
      res.push_back(i);
    }
  }
  cout << res.size() << endl;
  for (auto x : res) {
    cout << x << " ";
  }
}
