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: Time Limit Exceeded)

import math, string, itertools, fractions, heapq, collections, re,  array, bisect, sys, random, time, copy, functools, os
sys.setrecursionlimit(10**7)
inf = 10 ** 20
eps = 1.0 / 10**10
mod = 10**9+7
dd = [(-1, 0), (0, 1), (1, 0), (0, -1)]
ddn = [(-1, 0), (-1, 1), (0, 1), (1, 1), (1, 0), (1, -1), (0, -1), (-1, -1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def _I(): return int(sys.stdin.readline())
def _F(): return float(sys.stdin.readline())
def _pf(s): return print(s, flush=True)
def perr(s): return print(s, file=sys.stderr)
def debug(x, table):
    for name, val in table.items():
        if x is val:
            print('- {} -> {}'.format(name, val), file=sys.stderr)
            return None
N, M =LI()
AB = []
for i in range(N):
    AB.append(LI())
AB.sort(key=lambda x: x[0])
perr(AB)
#  AB_dict = {}
#  for ab in AB:
#      if ab[0] not in AB_dict.keys():
#          AB_dict[ab[0]] = [ab[1]]
#      else:
#          AB_dict[ab[0]].append(ab[1])
#  for k in AB_dict.keys():
#      AB_dict[k].sort()
#  perr(AB_dict)

ans = 0
tmp = []
crr = 0
for i in range(1, M+1):
    if crr < N:
        if AB[crr][0] == i:
            while True:
                if crr == N:
                    break
                if AB[crr][0] != i:
                    break
                tmp.append(AB[crr][1])
                crr += 1

    perr('tmp')
    perr(tmp)
    if len(tmp) == 0:
        continue

    max_v = max(tmp)
    ans += max_v
    tmp.remove(max_v)
    #  tmp.sort()
    #  ans+=tmp.pop()

"""
貪欲か？
日が長くて報酬の高いものから選んでいく
そうすると
10 1
9 10
みたいなときに10 1のほうから選んでしまうからだめ
残り日数を意識
残ってる中で最も報酬の高いものから？
2 10
1 11
のパターンで、先に1をやってしまうためだめ
2 10
1 11
1 15
のパターンで、1 1をやるのが正解
"""

#  ans = 0
#  tmp = []
#  for i in range(1, M+1):
#      # 残りi日
#      if i in AB_dict.keys():
#          tmp += AB_dict[i]
#      perr('tmp')
#      perr(tmp)
#      if len(tmp) == 0:
#          continue
#      tmp.sort()
#      #  max_v = max(tmp)
#      #  ans += max_v
#      ans += tmp.pop()
print(ans)