Problem:
Score : 400 points

Problem Statement

There are N one-off jobs available. If you take the i-th job and complete it, you will earn the reward of B_i after A_i days from the day you do it.

You can take and complete at most one of these jobs in a day.

However, you cannot retake a job that you have already done.

Find the maximum total reward that you can earn no later than M days from today.

You can already start working today.

Constraints

All values in input are integers.

1 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq A_i \leq 10^5

1 \leq B_i \leq 10^4

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
A_2 B_2
\vdots
A_N B_N

Output

Print the maximum total reward that you can earn no later than M days from today.

Sample Input 1

3 4
4 3
4 1
2 2

Sample Output 1

5

You can earn the total reward of 5 by taking the jobs as follows:

Take and complete the first job today. You will earn the reward of 3 after four days from today.

Take and complete the third job tomorrow. You will earn the reward of 2 after two days from tomorrow, that is, after three days from today.

Sample Input 2

5 3
1 2
1 3
1 4
2 1
2 3

Sample Output 2

10

Sample Input 3

1 1
2 1

Sample Output 3

0

Solution in Python (3.8.2) (judge verdict: Accepted)

import sys
from heapq import heappush,heappop

lines = []
for line in sys.stdin:
    lines.append(line.rstrip('\r\n'))
#print(lines)
n,m = lines[0].split()
n = int(n)
m = int(m)

a=[]
for i in range(n):
    x,y = lines[i+1].split()
    x = int(x)
    y = int(y)*(-1)
    a.append((x,y))
a.sort()
a.append((0,0))

h = []

cur = 0
summ = 0
for i in range(1,m+1):
    while(cur<n and a[cur][0]<=i):
        heappush(h,a[cur][1])
        cur+=1
    if(len(h)>0):
        summ+=heappop(h)*(-1)
print(summ)