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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
 
#define ms(s, n) memset(s, n, sizeof(s))
#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define FORd(i, a, b) for (int i = (a) - 1; i >= (b); i--)
#define FORall(it, a) for (__typeof((a).begin()) it = (a).begin(); it != (a).end(); it++)
#define sz(a) int((a).size())
#define present(t, x) (t.find(x) != t.end())
#define all(a) (a).begin(), (a).end()
#define uni(a) (a).erase(unique(all(a)), (a).end())
#define pb push_back
#define pf push_front
#define mp make_pair
#define fi first
#define se second
#define prec(n) fixed<<setprecision(n)
#define bit(n, i) (((n) >> (i)) & 1)
#define bitcount(n) __builtin_popcountll(n)
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int, int> pi;
typedef vector<int> vi;
typedef vector<pi> vii;
const int MOD = (int) 1e9 + 7;
const int MOD2 = 1007681537;
const int INF = (int) 1e9;
const ll LINF = (ll) 1e18;
const ld PI = acos((ld) -1);
const ld EPS = 1e-9;
inline ll gcd(ll a, ll b) {ll r; while (b) {r = a % b; a = b; b = r;} return a;}
inline ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
inline ll fpow(ll n, ll k, int p = MOD) {ll r = 1; for (; k; k >>= 1) {if (k & 1) r = r * n % p; n = n * n % p;} return r;}
template<class T> inline int chkmin(T& a, const T& val) {return val < a ? a = val, 1 : 0;}
template<class T> inline int chkmax(T& a, const T& val) {return a < val ? a = val, 1 : 0;}
inline ll isqrt(ll k) {ll r = sqrt(k) + 1; while (r * r > k) r--; return r;}
inline ll icbrt(ll k) {ll r = cbrt(k) + 1; while (r * r * r > k) r--; return r;}
inline void addmod(int& a, int val, int p = MOD) {if ((a = (a + val)) >= p) a -= p;}
inline void submod(int& a, int val, int p = MOD) {if ((a = (a - val)) < 0) a += p;}
inline int mult(int a, int b, int p = MOD) {return (ll) a * b % p;}
inline int inv(int a, int p = MOD) {return fpow(a, p - 2, p);}
inline int sign(ld x) {return x < -EPS ? -1 : x > +EPS;}
inline int sign(ld x, ld y) {return sign(x - y);}
#define db(x) cerr << #x << " = " << (x) << " ";
#define endln cerr << "\n";
 
template<class T, T oo> struct Dinic {
    static const int MAXV = 2e7 + 5;
    static const int MAXE = 2e7 + 5;
    int n, s, t, E;
    int adj[MAXE], nxt[MAXE], lst[MAXV], ptr[MAXV], lev[MAXV], que[MAXV];
    T flw[MAXE], cap[MAXE];
    void init(int n, int s, int t) {
        this->n = n, this->s = s, this->t = t, E = 0;
        fill_n(lst, n, -1);
    }
    void add(int u, int v, T c1, T c2) {
        while (n <= max(u, v)) {
            lst[n++] = -1;
        }
        adj[E] = v, flw[E] = 0, cap[E] = c1, nxt[E] = lst[u], lst[u] = E++;
        adj[E] = u, flw[E] = 0, cap[E] = c2, nxt[E] = lst[v], lst[v] = E++;
    }
    int bfs() {
        fill_n(lev, n, -1), lev[s] = 0;
        int qsize = 0;
        que[qsize++] = s;
        for (int i = 0; i < qsize; i++) {
            for (int u = que[i], e = lst[u]; ~e; e = nxt[e]) {
                int v = adj[e];
                if (flw[e] < cap[e] && !~lev[v]) {
                    lev[v] = lev[u] + 1;
                    que[qsize++] = v;
                }
            }
        }
        return lev[t] != -1;
    }
    T dfs(int u, T bot) {
        if (u == t) return bot;
        for (int& e = ptr[u]; ~e; e = nxt[e]) {
            int v = adj[e];
            T delta = 0;
            if (lev[v] == lev[u] + 1 && flw[e] < cap[e] && (delta = dfs(v, min(bot, cap[e] - flw[e]))) > 0) {
                flw[e] += delta; flw[e ^ 1] -= delta;
                return delta;
            }
        }
        return 0;
    }
    T maxflow() {
        T total = 0;
        while (bfs()) {
            for (int i = 0; i < n; i++) ptr[i] = lst[i];
            for (T delta = dfs(s, oo); delta > 0; delta = dfs(s, oo)) total += delta;
        }
        return total;
    }
};
Dinic<int, (int) 1e9> dinic;
 
const int maxn = 1e6 + 5;
int n, m;
int l[maxn];
int r[maxn];
int tot;
int len[maxn];
 
struct SegTree {
    int st[maxn << 2];
//    void build(int p, int L, int R) {
//        st[p] = tot++;
//        if (L == R) {
//            dinic.add(st[p], dinic.t, 1, 0);
//            return;
//        }
//        build(p << 1, L, L + R >> 1);
//        build(p << 1 | 1, (L + R >> 1) + 1, R);
//        dinic.add(st[p], st[p << 1], INF, 0);
//        dinic.add(st[p], st[p << 1 | 1], INF, 0);
//    }
    void upd(int p, int l, int r, int L, int R, int ix) {
        if (!st[p]) {
            len[p] = R - L + 1;
            st[p] = tot++;
            if (p > 1) {
                dinic.add(st[p >> 1], st[p], R - L + 1, 0);
            }
        }
        if (l > R || r < L) return;
        if (l <= L && r >= R) {
            dinic.add(ix, st[p], 1, 0);
            return;
        }
        upd(p << 1, l, r, L, L + R >> 1, ix);
        upd(p << 1 | 1, l, r, (L + R >> 1) + 1, R, ix);
    }
} seg;
 
void work(int l, int r, int ix) {
    seg.upd(1, l, r, 1, m, ix);
}
 
void solve() {
    cin >> n >> m;
    tot = n;
    int s = tot++, t = tot++;
    dinic.init(tot, s, t);
    FOR(i, 0, n) {
        dinic.add(s, i, 1, 0);
    }
//    seg.build(1, 1, m);
    FOR(i, 0, n) {
        cin >> l[i] >> r[i];
        work(1, l[i], i);
        work(r[i], m, i);
    }
    FOR(p, 0, maxn << 2) if (seg.st[p] && (!seg.st[p << 1])) {
        dinic.add(seg.st[p], dinic.t, len[p], 0);
    }
    cout << n - dinic.maxflow() << "\n";
}
 
int main() {
    int JUDGE_ONLINE = 1;
    if (fopen("in.txt", "r")) {
        JUDGE_ONLINE = 0;
        assert(freopen("in.txt", "r", stdin));
        //assert(freopen("out.txt", "w", stdout));
    }
    else {
        ios_base::sync_with_stdio(0), cin.tie(0);
    }
    solve();
    if (!JUDGE_ONLINE) {
        //cout << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
    }
    return 0;
}