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-----
4 4
0 3
2 3
1 3
3 4

-----Sample Output-----
0

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

Difficulty: competition

Starter code:

Solutions:
import sys

input=sys.stdin.readline

N,M=list(map(int,input().split()))

# N: 処理する区間の長さ
INF = 2**31-1

LV = (M+2-1).bit_length()
N0 = 2**LV
data = [0]*(2*N0)
lazy = [0]*(2*N0)

def gindex(l, r):
    L = (l + N0) >> 1; R = (r + N0) >> 1
    lc = 0 if l & 1 else (L & -L).bit_length()
    rc = 0 if r & 1 else (R & -R).bit_length()
    for i in range(LV):
        if rc <= i:
            yield R
        if L < R and lc <= i:
            yield L
        L >>= 1; R >>= 1

# 遅延伝搬処理
def propagates(*ids):
    for i in reversed(ids):
        v = lazy[i-1]
        if not v:
            continue
        lazy[2*i-1] += v; lazy[2*i] += v
        data[2*i-1] += v; data[2*i] += v
        lazy[i-1] = 0

# 区間[l, r)にxを加算
def update(l, r, x):
    *ids, = gindex(l, r)
    propagates(*ids)

    L = N0 + l; R = N0 + r
    while L < R:
        if R & 1:
            R -= 1
            lazy[R-1] += x; data[R-1] += x
        if L & 1:
            lazy[L-1] += x; data[L-1] += x
            L += 1
        L >>= 1; R >>= 1
    for i in ids:
        data[i-1] = min(data[2*i-1], data[2*i])

# 区間[l, r)内の最小値を求める
def query(l, r):
    propagates(*gindex(l, r))
    L = N0 + l; R = N0 + r

    s = INF
    while L < R:
        if R & 1:
            R -= 1
            s = min(s, data[R-1])
        if L & 1:
            s = min(s, data[L-1])
            L += 1
        L >>= 1; R >>= 1
    return s

for i in range(1,M+1):
    update(0,i+1,1)

add=M-N
hito=[]
for i in range(N):
    L,R=list(map(int,input().split()))
    hito.append((L,R))
hito.sort()
#test=[query(i,i+1) for i in range(M+2)]
#print(test)
for l,r in hito:
    update(0,r+1,-1)
    #test=[query(i,i+1) for i in range(M+2)]
    #print(test)
    m=query(l+1,M+2)+l
    add=min(m,add)

print((max(-add,0)))


from operator import itemgetter
from heapq import heappush, heappop
N, M = list(map(int, input().split()))
LR = [list(map(int, input().split())) for _ in range(N)]
LR.sort(key=itemgetter(1))
A = []
ans_left = 0
idx = 1
for _, r in LR:
    if r==M+1 or idx==M+1:
        break
    ans_left += 1
    idx = max(idx+1, r+1)

idx_LR = 0
q = []
for i in range(M+1-ans_left, M+1):
    while idx_LR<N and LR[idx_LR][1]<=i:
        l, _ = LR[idx_LR]
        heappush(q, l)
        idx_LR += 1
    heappop(q)
while idx_LR<N:
    l, _ = LR[idx_LR]
    q.append(l)
    idx_LR += 1

q.sort(reverse=True)
ans_right = 0
idx = M
for l in q:
    if l==0 or idx==0 or ans_right+ans_left==M:
        break
    ans_right += 1
    idx = min(l-1, idx-1)
ans = N - ans_left - ans_right
print(ans)