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: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define R register int
#define AC 400100
#define ac 1601000//error!!!数据范围是2 00000, 不是1开头！！！

int n, m, ans = -INT_MAX, w;
int Head[AC], Next[ac], date[ac], tot;
int tree[ac], lazy[ac], l[ac], r[ac], l_[AC], r_[AC];

inline int read()
{
	int x = 0;char c = getchar();
	while(c > '9' || c < '0') c = getchar();
	while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
	return x;
}

inline void add(int f, int w)
{
	date[++tot] = w, Next[tot] = Head[f], Head[f] = tot;
}

inline void upmax(int &a, int b)
{
	if(b > a) a = b;
}

void pre()
{
	n = read(), m = read();
	for(R i = 1; i <= n; i ++) 
		l_[i] = read(), r_[i] = read(), add(r_[i], i);
}

inline void pushdown(int x)
{
	if(lazy[x])
	{
		int ll = x * 2, rr = ll + 1;
		lazy[ll] += lazy[x], lazy[rr] += lazy[x];
		tree[ll] += lazy[x], tree[rr] += lazy[x];//这里因为是+=，所以必须用lazy[x]，不然会将lazy[ll]中的一些东西重复统计
		lazy[x] = 0;//最后才清空！！！！！！！！！！
	}//error!!!是区间加，不是赋值，不能直接覆盖，要+=
}

inline void update(int x)
{
	tree[x] = max(tree[x * 2], tree[x * 2 + 1]); 
}

void build(int x, int ll, int rr)
{
	l[x] = ll, r[x] = rr;
	if(ll == rr) 
	{
		tree[x] = -ll + 1;
		return ;
	}
	int mid = (ll + rr) >> 1;
	build(x * 2, ll, mid);
	build(x * 2 + 1, mid + 1, rr);
	update(x);
}

void change(int x, int ll, int rr)
{
	pushdown(x);
	if(l[x] == ll && r[x] == rr)
	{
		lazy[x] += w, tree[x] += w;
		return ;
	}
	int mid = (l[x] + r[x]) >> 1;
	if(rr <= mid) change(x * 2, ll, rr);
	else if(ll > mid) change(x * 2 + 1, ll, rr);
	else
	{
		change(x * 2, ll, mid);
		change(x * 2 + 1, mid + 1, rr);
	}
	update(x);
}
	
void find(int x, int ll, int rr)
{
	pushdown(x);
	if(l[x] == ll && r[x] == rr) 
	{
		upmax(ans, tree[x]);
		return ;	
	}
	int mid = (l[x] + r[x]) >> 1;
	if(rr <= mid) find(x * 2, ll, rr);
	else if(ll > mid) find(x * 2 + 1, ll, rr);
	else find(x * 2, ll, mid), find(x * 2 + 1, mid + 1, rr);//这是取max啊，，，，
}

void work()
{
	int now;
	ans = n - m;//r = 0的情况
	for(R i = Head[m + 1]; i; i = Next[i])
	{
		now = date[i], w = 1;
		change(1, l_[now] + 1, m + 1);
		//find(1, 4, 4);
	}
	//find(1, 4, 4);
	upmax(ans, tree[1]);
	for(R i = m; i; -- i)
	{
		w = -1;
		change(1, 1, i + 1);
		for(R j = Head[i]; j; j = Next[j])
		{
			now = date[j], w = 1;
			change(1, l_[now] + 1, i + 1);
		}
		find(1, 1, i + 1);//左端点在后面就不合法了
	}
	printf("%d\n", ans);
}

int main()
{
	pre();
	build(1, 1, m + 1);//要多出一位来代表左端点取0的情况
	work();//ri最大居然可以到m+1...
	return 0;
}