Problem:
Score : 1000 points

Problem Statement

There are M chairs arranged in a line. The coordinate of the i-th chair (1 ≤ i ≤ M) is i.

N people of the Takahashi clan played too much games, and they are all suffering from backaches. They need to sit in chairs and rest, but they are particular about which chairs they sit in. Specifically, the i-th person wishes to sit in a chair whose coordinate is not greater than L_i, or not less than R_i. Naturally, only one person can sit in the same chair.

It may not be possible for all of them to sit in their favorite chairs, if nothing is done.
Aoki, who cares for the health of the people of the Takahashi clan, decides to provide additional chairs so that all of them can sit in chairs at their favorite positions.

Additional chairs can be placed at arbitrary real coordinates. Find the minimum required number of additional chairs.

Constraints

1 ≤ N,M ≤ 2 × 10^5

0 ≤ L_i < R_i ≤ M + 1(1 ≤ i ≤ N)

All input values are integers.

Input

Input is given from Standard Input in the following format:

N M
L_1 R_1
:
L_N R_N

Output

Print the minimum required number of additional chairs.

Sample Input 1

4 4
0 3
2 3
1 3
3 4

Sample Output 1

0

The four people can sit in chairs at the coordinates 3, 2, 1 and 4, respectively, and no more chair is needed.

Sample Input 2

7 6
0 7
1 5
3 6
2 7
1 6
2 6
3 7

Sample Output 2

2

If we place additional chairs at the coordinates 0 and 2.5, the seven people can sit at coordinates 0, 5, 3, 2, 6, 1 and 2.5, respectively.

Sample Input 3

3 1
1 2
1 2
1 2

Sample Output 3

2

Sample Input 4

6 6
1 6
1 6
1 5
1 5
2 6
2 6

Sample Output 4

2

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
// #define int long long
// #define double long double
#define rep(i,n) for (int i=0; i<(int)(n); ++i)
#define rep1(i,n) for (int i=1; i<(int)(n); ++i)
#define repeq(i,n) for (int i=0; i<=(int)(n); ++i)
#define rep1eq(i,n) for (int i=1; i<=(int)(n); ++i)
#define rrep(i,n) for (int i=(int)(n)-1; i>=0; --i)
#define rrep1(i,n) for (int i=(int)(n)-1; i>0; --i)
#define rrepeq(i,n) for (int i=(int)(n); i>=0; --i)
#define rrep1eq(i,n) for (int i=(int)(n); i>0; --i)
#define REP(i,a,b) for (int i=(int)(a); i<=(int)(b); ++i)
#define RREP(i,a,b) for (int i=(int)(a); i>=(int)(b); --i)
#define ALL(a) (a).begin(), (a).end()
#define RALL(a) (a).rbegin(), (a).rend()
using ll = long long;
using vi = vector<int>;
using vl = vector<ll>;
using vb = vector<bool>;
template<typename T> using Graph = vector<vector<T>>;
template<typename T> using Spacial = vector<vector<vector<T>>>;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
const int MOD = 1e9+7;
const int MOD2 = 998244353;
// const double EPS = 1e-9;
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
string interval[2] = {" ", "\n"}; // {" ", "\n"}

template<typename T> struct is_plural : false_type{};
template<typename T1, typename T2> struct is_plural<pair<T1, T2>> : true_type{};
template<typename T> struct is_plural<vector<T>> : true_type{};
template<typename T> struct is_plural<complex<T>> : true_type{};
 
template<typename T1, typename T2> istream &operator>>(istream &is, pair<T1, T2> &p) { return is >> p.first >> p.second; }
template<typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &p) { return os << p.first << " " << p.second; }
template<typename T> istream &operator>>(istream &is, vector<T> &vec) { for (auto itr = vec.begin(); itr != vec.end(); ++itr) is >> *itr; return is; }
template<typename T> ostream &operator<<(ostream &os, const vector<T> &vec) { if (vec.empty()) return os; bool pl = is_plural<T>(); os << vec.front(); for (auto itr = ++vec.begin(); itr != vec.end(); ++itr) os << interval[pl] << *itr; return os; }
 
bool CoutYN(bool a, string y = "Yes", string n = "No") { cout << (a ? y : n) << "\n"; return a; }

template<typename T1, typename T2> inline bool chmax(T1 &a, T2 b) { return a < b && (a = b, true); }
template<typename T1, typename T2> inline bool chmin(T1 &a, T2 b) { return a > b && (a = b, true); }

long long modpow(int a, long long n, int mod = MOD) { long long ret = 1; do { if (n & 1) ret = ret * a % mod; a = 1LL * a * a % mod; } while (n >>= 1); return ret; }

template<typename T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; }
template<typename T> T LCM(T a, T b) { return a / GCD(a, b) * b; }

template<typename T1, typename T2> bool CompareBySecond(pair<T1, T2> a, pair<T1, T2> b) { return a.second != b.second ? a.second < b.second : a.first < b.first; }
template<typename T1, typename T2> bool CompareByInverse(pair<T1, T2> a, pair<T1, T2> b) { return a.first != b.first ? a.first < b.first : a.second > b.second; }


/* -------- <templates end> -------- */


template<typename Monoid>
struct SegmentTree {
  using F = function<Monoid(Monoid, Monoid)>;

  int sz;
  vector<Monoid> seg;

  const F f;
  const Monoid M1;

  SegmentTree(int n, const F f, const Monoid &M1) : f(f), M1(M1) {
    sz = 1;
    while (sz < n) sz <<= 1;
    seg.assign(2*sz, M1);
  }

  void set(int k, const Monoid &x) {
    seg[k+sz] = x;
  }

  void build() {
    for (int k = sz-1; k>0; --k) {
      seg[k] = f(seg[2*k+0], seg[2*k+1]);
    }
  }

  void update(int k, const Monoid &x) {
    k += sz;
    seg[k] = x;
    while (k >>= 1) {
      seg[k] = f(seg[2*k+0], seg[2*k+1]);
    }
  }

// [a, b) 0-indexed
  Monoid query(int a, int b) {
    Monoid L = M1, R = M1;
    for (a += sz, b += sz; a < b; a >>= 1, b >>= 1) {
      if (a & 1) L = f(L, seg[a++]);
      if (b & 1) R = f(seg[--b], R);
    }
    return f(L, R);
  }

  Monoid operator[](const int &k) const {
    return seg[k+sz];
  }

  template<typename C>
  int find_subtree(int a, const C &check, Monoid &M, bool type) {
    while (a < sz) {
      Monoid nxt = type ? f(seg[2*a+type], M) : f(M, seg[2*a+type]);
      if (check(nxt)) a = 2*a + type;
      else M = nxt, a = 2*a+1 - type;
    }
    return a-sz;
  }

// [a, x] fills check
  template<typename C>
  int find_first(int a, const C &check) {
    Monoid L = M1;
    if (a <= 0) {
      if (check(f(L, seg[1]))) return find_subtree(1, check, L, false);
      return -1;
    }
    int b = sz;
    for (a += sz, b += sz; a < b; a >>= 1, b >>= 1) {
      if (a & 1) {
        Monoid nxt = f(L, seg[a]);
        if (check(nxt)) return find_subtree(a, check, L, false);
        L = nxt;
        ++a;
      }
    }
    return -1;
  }

// [x, b) fills check
  template<typename C>
  int find_last(int b, const C &check) {
    Monoid R = M1;
    if (b >= sz) {
      if (check(f(seg[1], R))) return find_subtree(1, check, R, true);
      return -1;
    }
    int a = sz;
    for (b += sz; a < b; a >>= 1, b >>= 1) {
      if (b & 1) {
        Monoid nxt = f(seg[--b], R);
        if (check(nxt)) return find_subtree(b, check, R, true);
        R = nxt;
      }
    }
    return -1;
  }
};

void solve() {
  int n, m; cin >> n >> m;
  vector<pii> g; g.reserve(n);
  rep(i,n) {
    int l, r; cin >> l >> r; --l, --r;
    g.emplace_back(l, r);
  }

  sort(ALL(g), CompareByInverse<int, int>);

  SegmentTree<bool> seg(m, [](bool a, bool b) { return a | b; }, false);
  rep(i,m) seg.set(i, true);
  seg.build();

  int add = 0;

  for (auto &p : g) {
    int r = seg.find_first(p.second, [](bool b) { return b; });
    if (r != -1) {
      seg.update(r, false);
      continue;
    }

    int l = seg.find_last(p.first + 1, [](bool b) { return b; });
    if (l != -1) {
      seg.update(l, false);
      continue;
    }

    ++add;
  }

  cout << add << endl;
}


/* -------- <programs end> -------- */


signed main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(10);
  solve();
  return 0;
}