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)

'''
rows , days = map(int,input().split())
total = 0
arr2 = [[] for _ in range(days)]

for i in range(rows):
    day , r = list(map(int,input().split()))
    if day <= days:
        arr2[day].append(r)

print(arr2)
'''

#標準入力受け取り
N, M = list(map(int, input().split()))
jobs = [[] for _ in range(M)]

for i in range(N):
    d, r = list(map(int, input().split()))
    if d-1 < M:
        jobs[d-1].append(r)

print(jobs)
#解法
rewards = 0
li = []
for i in range(M):
    #候補をリストに追加(給料が間に合うもの仕事が一日ごとに増える)
    li.extend(jobs[i])
    if len(li) >0:
        #最大値を求める(給料が高いバイトをしたい)
        reward = max(li)
        rewards += reward
        #使った値は取り除く(同じバイトはNGのため)
        li.remove(reward)

print(rewards)