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>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
namespace Dup4 {
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar()) f ^= (c == '-');
  for (; isdigit(c); c = getchar()) x = x * 10 + (c - '0');
  return x * (f ? 1 : -1);
}
template <typename T>
inline void read(T &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar()) f ^= (c == '-');
  for (; isdigit(c); c = getchar()) x = x * 10 + (c - '0');
  x *= f ? 1 : -1;
}
template <typename T>
inline void out(T x) {
  if (!x) {
    putchar(48);
    return;
  }
  if (x < 0) x = -x, putchar('-');
  int bit[20], i, p = 0;
  for (; x; x /= 10) bit[++p] = x % 10;
  for (i = p; i; --i) putchar(bit[i] + 48);
}
template <typename T>
inline void output(T x) {
  out(x), putchar('\n');
}
template <typename T>
inline T gcd(T a, T b) {
  while (b ^= a ^= b ^= a %= b)
    ;
  return a;
}
const double PI = acos(-1.0);
const double EI = exp(1.0);
const double eps = 1e-8;
const int INF = 0x3f3f3f3f;
const long long INFLL = 0x3f3f3f3f3f3f3f3fll;
}  // namespace Dup4
using namespace Dup4;
namespace FastIO {
bool IOerror = false;
inline char NC() {
  static char buf[10000005], *p1 = buf + 10000005, *pend = buf + 10000005;
  if (p1 == pend) {
    p1 = buf;
    pend = buf + fread(buf, 1, 10000005, stdin);
    if (pend == p1) {
      IOerror = true;
      return -1;
    }
  }
  return *p1++;
}
inline bool blank(char ch) {
  return ch == ' ' || ch == '\n' || ch == '\r' || ch == '\t';
}
inline void __read(char &x) {
  char ch;
  while (blank(ch = NC()))
    ;
  if (IOerror) {
    x = -1;
    return;
  }
  x = ch;
}
template <typename T>
inline void __read(T &x) {
  char ch;
  while (blank(ch = NC()))
    ;
  if (IOerror) {
    x = -1;
    return;
  }
  bool flag = false;
  if (ch == '-') {
    flag = true;
    ch = NC();
  }
  if (!isdigit(ch))
    while (!isdigit(ch = NC()))
      ;
  for (x = ch - '0'; isdigit(ch = NC()); x = x * 10 + ch - '0')
    ;
  if (flag) x *= -1;
}
}  // namespace FastIO
using namespace FastIO;
const int maxn = (int)1e2 + 10;
const int Maxn = (int)2e6 + 10;
const int MOD = (int)1e9 + 7;
int n, m, ans;
int used[maxn];
inline bool Input() {
  if (__read(n), __read(m), n == EOF) return false;
  ans = m;
  int l, r;
  memset(used, (false), sizeof(used));
  ;
  for (int i = (1); i <= (n); ++i) {
    __read(l), __read(r);
    for (int j = (l); j <= (r); ++j)
      if (used[j] == false) {
        ans--;
        used[j] = true;
      }
  }
  return true;
}
inline void Solve() {
  output(ans);
  for (int i = 1, pos = 0; i <= m; i++)
    if (used[i] == false) {
      if (pos) putchar(' ');
      out(i);
      pos++;
    }
  putchar('\n');
}
int main() {
  while (Input()) Solve();
  return 0;
}