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 (3.4.3) (judge verdict: Wrong Answer)

import heapq
n,m = map(int,input().split())
j = []
for i in range(n):
    j1 = list(map(int,input().split()))
    j.append(j1)
j.sort(key=lambda x:x[1],reverse=True)
j.sort(key=lambda x:x[0])
day = 1
i = 0
q = []
ans = 0
count = 0
while day <= m and i < n:
    if day == j[i][0]:
        heapq.heappush(q,-(j[i][1]))
        ans = ans + heapq.heappop(q)*(-1)
        day = day + 1
        i = i + 1
    elif day > j[i][0]:
        heapq.heappush(q,-(j[i][1]))
        i = i + 1
        count = count + 1
    else:
        day = day + 1
while day <= m and count > 0:
        ans = ans + heapq.heappop(q)*(-1)
        day = day + 1
        count = count - 1
print(ans)