A social network for dogs called DH (DogHouse) has k special servers to recompress uploaded videos of cute cats. After each video is uploaded, it should be recompressed on one (any) of the servers, and only after that it can be saved in the social network.

We know that each server takes one second to recompress a one minute fragment. Thus, any server takes m seconds to recompress a m minute video.

We know the time when each of the n videos were uploaded to the network (in seconds starting from the moment all servers started working). All videos appear at different moments of time and they are recompressed in the order they appear. If some video appeared at time s, then its recompressing can start at that very moment, immediately. Some videos can await recompressing when all the servers are busy. In this case, as soon as a server is available, it immediately starts recompressing another video. The videos that await recompressing go in a queue. If by the moment the videos started being recompressed some servers are available, then any of them starts recompressing the video.

For each video find the moment it stops being recompressed.


-----Input-----

The first line of the input contains integers n and k (1 ≤ n, k ≤ 5·10^5) — the number of videos and servers, respectively.

Next n lines contain the descriptions of the videos as pairs of integers s_{i}, m_{i} (1 ≤ s_{i}, m_{i} ≤ 10^9), where s_{i} is the time in seconds when the i-th video appeared and m_{i} is its duration in minutes. It is guaranteed that all the s_{i}'s are distinct and the videos are given in the chronological order of upload, that is in the order of increasing s_{i}.


-----Output-----

Print n numbers e_1, e_2, ..., e_{n}, where e_{i} is the time in seconds after the servers start working, when the i-th video will be recompressed.


-----Examples-----
Input
3 2
1 5
2 5
3 5

Output
6
7
11

Input
6 1
1 1000000000
2 1000000000
3 1000000000
4 1000000000
5 1000000000
6 3

Output
1000000001
2000000001
3000000001
4000000001
5000000001
5000000004

Source: codeforces | Difficulty: MEDIUM_HARD
Tags: Data structures

Implementation


Solutions:
from heapq import heappop, heappush
from sys import stdin, stdout
(read, read_array) = (stdin.readline, lambda : stdin.readline().split())
write = lambda *args, **kw: stdout.write(kw.get('sep', ' ').join((str(a) for a in args)) + kw.get('end', '\n'))
write_array = lambda arr, **kw: stdout.write(kw.get('sep', ' ').join((str(a) for a in arr)) + kw.get('end', '\n'))
(read_int, read_int_array) = (lambda : int(read()), lambda : [int(p) for p in read_array()])
(read_float, read_float_array) = (lambda : float(read()), lambda : [float(p) for p in read_array()])
(n, k) = read_int_array()
heap = []
busy = 0
time = 0
finish = [0] * n
for i in range(n):
	if busy == k:
		time = heappop(heap)
		busy -= 1
	else:
		time = 0
	(start, minutes) = read_int_array()
	if start > time:
		time = start
	heappush(heap, time + minutes)
	finish[i] = time + minutes
	busy += 1
write_array(finish, sep='\n')


import heapq
from sys import stdin, stdout

def __main__(n, k):
	servers = [0] * k
	times = []
	for i in range(n):
		(s, m) = list(map(int, stdin.readline().split()))
		time = max(servers[0], s)
		heapq.heapreplace(servers, time + m)
		times.append(time + m)
	stdout.write('\n'.join((str(time) for time in times)) + '\n')
(n, k) = list(map(int, stdin.readline().split()))
__main__(n, k)


import heapq

def __main__(n, k):
	servers = [0] * k
	times = []
	for i in range(n):
		(s, m) = list(map(int, input().split()))
		time = max(servers[0], s)
		heapq.heapreplace(servers, time + m)
		times.append(time + m)
	print('\n'.join((str(time) for time in times)))
(n, k) = list(map(int, input().split()))
__main__(n, k)


from sys import stdin, stdout
import heapq
lines = stdin.readlines()
(n, k) = (int(lines[0].split()[0], 10), int(lines[0].split()[1], 10))
a = [int(x.split()[0], 10) for x in lines[1:]]
b = [int(x.split()[1], 10) for x in lines[1:]]
heap = [a[i] + b[i] for i in range(k)]
answers = heap[:]
heapq.heapify(heap)
for i in range(k, n):
	will_load = int(max(int(heapq.heappop(heap)), a[i]) + b[i])
	heapq.heappush(heap, will_load)
	answers.append(will_load)
stdout.write('\n'.join([str(x) for x in answers]))


import heapq
f = lambda : map(int, input().split())
(n, k) = f()
p = [0] * k
q = [0] * n
for i in range(n):
	(s, m) = f()
	q[i] = max(p[0], s) + m
	heapq.heapreplace(p, q[i])
print('\n'.join(map(str, q)))


import atexit
import io
import sys
_INPUT_LINES = sys.stdin.read().splitlines()
input = iter(_INPUT_LINES).__next__
_OUTPUT_BUFFER = io.StringIO()
sys.stdout = _OUTPUT_BUFFER

@atexit.register
def write():
	sys.__stdout__.write(_OUTPUT_BUFFER.getvalue())

def main():
	import heapq
	(n, k) = [int(i) for i in input().split()]
	(s, m) = ([0] * n, [0] * n)
	for i in range(n):
		(s[i], m[i]) = [int(i) for i in input().split()]
	current_time = 0
	event = []
	ans = [0] * n
	i = 0
	while len(event) or i < n:
		if len(event) == 0:
			current_time = max(current_time, s[i])
			heapq.heappush(event, (current_time + m[i], i))
			i += 1
			continue
		(e_time, e_id) = event[0]
		if e_time == current_time or i == n:
			current_time = max(current_time, e_time)
			ans[e_id] = e_time
			heapq.heappop(event)
			continue
		if s[i] < e_time and len(event) < k:
			current_time = max(current_time, s[i])
			heapq.heappush(event, (current_time + m[i], i))
			i += 1
			continue
		else:
			current_time = ans[e_id] = e_time
			heapq.heappop(event)
			continue
	print('\n'.join([str(i) for i in ans]))
main()


import sys
input = sys.stdin.readline
rInt = lambda : int(input())
mInt = lambda : map(int, input().split())
rList = lambda : list(map(int, input().split()))
(n, k) = mInt()
inp = []
for i in range(n):
	inp.append(rList())
avail = k
from heapq import *
events = []
from collections import deque
queue = deque()
out = [-1] * n
heappush(events, (inp[0][0], 1, 0))
while events:
	nex = heappop(events)
	if nex[1]:
		queue.append(nex[2])
		if nex[2] < n - 1:
			heappush(events, (inp[nex[2] + 1][0], 1, nex[2] + 1))
	else:
		avail += 1
	t = nex[0]
	while avail and queue:
		avail -= 1
		nex = queue.popleft()
		end = t + inp[nex][1]
		out[nex] = end
		heappush(events, (end, 0))
print(*out, sep='\n')


from sys import stdin, stdout
lines = stdin.readlines()
from random import random
from time import time
t1 = time()
(n, k) = (int(lines[0].split()[0]), int(lines[0].split()[1]))
a = [int(x.split()[0]) for x in lines[1:]]
b = [int(x.split()[1]) for x in lines[1:]]
import heapq
heap = []
free_servers = k
answers = []
heap = [a[i] + b[i] for i in range(k)]
answers = heap[:]
heapq.heapify(heap)
for i in range(k, n):
	cur_min = int(heapq.heappop(heap))
	will_load = int(max(cur_min, a[i]) + b[i])
	heapq.heappush(heap, will_load)
	answers.append(will_load)
t2 = time()
stdout.write('\n'.join([str(x) for x in answers]))


import heapq
(n, k) = map(int, input().split())
pq = [0] * k
res = [None] * n
for i in range(n):
	(s, m) = map(int, input().split())
	time = max(pq[0], s) + m
	heapq.heapreplace(pq, time)
	res[i] = time
print('\n'.join(map(str, res)))