Code Forces 644B Processing Queries

B. Processing Queries

time limit per test5 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard output

In this problem you have to simulate the workflow of one-thread server. There are n queries to process, the i-th will be received at moment ti and needs to be processed for di units of time. All ti are guaranteed to be distinct.

When a query appears server may react in three possible ways:

If server is free and query queue is empty, then server immediately starts to process this query.

If server is busy and there are less than b queries in the queue, then new query is added to the end of the queue.

If server is busy and there are already b queries pending in the queue, then new query is just rejected and will never be processed.

As soon as server finished to process some query, it picks new one from the queue (if it’s not empty, of course). If a new query comes at some moment x, and the server finishes to process another query at exactly the same moment, we consider that first query is picked from the queue and only then new query appears.

For each query find the moment when the server will finish to process it or print -1 if this query will be rejected.

Input

The first line of the input contains two integers n and b (1 ≤ n, b ≤ 200 000) — the number of queries and the maximum possible size of the query queue.

Then follow n lines with queries descriptions (in chronological order). Each description consists of two integers ti and di (1 ≤ ti, di ≤ 109), where ti is the moment of time when the i-th query appears and di is the time server needs to process it. It is guaranteed that ti - 1 < ti for all i > 1.

Output

Print the sequence of n integers e1, e2, …, en, where ei is the moment the server will finish to process the i-th query (queries are numbered in the order they appear in the input) or  - 1 if the corresponding query will be rejected.

Examples

input

5 1

2 9

4 8

10 9

15 2

19 1

output

11 19 -1 21 22

input

4 1

2 8

4 8

10 9

15 2

output

10 18 27 -1

#include <iostream>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <queue>

using namespace std;
#define MAX 200000
struct Node
{
    int id;
    long long int st;
    long long int time;
}a[MAX+5];
int cmp(Node a,Node b)
{
    return a.st<b.st;
}
queue<Node> q;
long long int ans[MAX+5];
int n,b;
int main()
{
    scanf("%d%d",&n,&b);
    for(int i=1;i<=n;i++)
    {
        scanf("%lld%lld",&a[i].st,&a[i].time);
        a[i].id=i;
    }
    sort(a+1,a+n+1,cmp);
    int i=2;
    long long int now=a[1].st+a[1].time;
    ans[1]=now;
    while(1)
    {
        if(i>n&&q.size()==0)
            break;
        while(i<=n)
        {
            if(a[i].st>=now)
                break;
            if(q.size()>=b)
                ans[a[i++].id]=-1;
            else
            {
                q.push(a[i++]);
            }
        }
        if(q.size()==0)
        {now=a[i].st+a[i].time;ans[i]=now;i++;continue;}
        Node term=q.front();
        q.pop();
        now+=term.time;
        ans[term.id]=now;
    }
    for(int i=1;i<=n;i++)
    {
        if(i==n)
            printf("%lld\n",ans[i]);
        else
            printf("%lld ",ans[i]);
    }
    return 0;
}