Problem:
There are n processes in a queue. Each process has namei and timei. The round-robin scheduling handles the processes in order. A round-robin scheduler gives each process a quantum (a time slot) and interrupts the process if it is not completed by then. The process is resumed and moved to the end of the queue, then the scheduler handles the next process in the queue.

For example, we have the following queue with the quantum of 100ms.

A(150) - B(80) - C(200) - D(200)

First, process A is handled for 100ms, then the process is moved to the end of the queue with the remaining time (50ms).

B(80) - C(200) - D(200) - A(50)

Next, process B is handled for 80ms. The process is completed with the time stamp of 180ms and removed from the queue.

C(200) - D(200) - A(50)

Your task is to write a program which simulates the round-robin scheduling.

Input

n q

name1 time1

name2 time2

...

namen timen

In the first line the number of processes n and the quantum q are given separated by a single space.

In the following n lines, names and times for the n processes are given. namei and timei are separated by a single space.

Output

For each process, prints its name and the time the process finished in order.

Constraints

1 ≤ n ≤ 100000

1 ≤ q ≤ 1000

1 ≤ timei ≤ 50000

1 ≤ length of namei ≤ 10

1 ≤ Sum of timei ≤ 1000000

Sample Input 1

5 100
p1 150
p2 80
p3 200
p4 350
p5 20

Sample Output 1

p2 180
p5 400
p1 450
p3 550
p4 800

Notes

Template in C

Solution in C (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define NAMELENGTH 11
//#define MAXSIZE 100001

struct ALDS1_3_C{
	
	char name[NAMELENGTH];
	int time;
	
};
typedef struct ALDS1_3_C ALDS;

int front = 0, rear = 0;

void enqueue(ALDS queue[], ALDS data, int n);
ALDS dequeue(ALDS queue[], int idx, int n);

void display(ALDS data[], int from, int to);

int main(void){
	
	int n, q;
	int i;
	int size;
	int doneTime = 0;
	char dummy;
	ALDS temp;
	ALDS *queue = NULL;
	
	//プロセス数nとクオンタムqを入力
	scanf("%d %d", &n, &q);
	
	//初期サイズはn, rearもn
	size = n;
	rear = n;
	
	//nに応じた大きさをqueueに取得
	queue = (ALDS*)malloc((sizeof(ALDS)*n*2)+1);
	
	
	//入力開始
	for(i=0; i<n; i++){
		
		scanf("%s%c%d%c", queue[i].name, &dummy, &queue[i].time, &dummy);
		
	}
	
	
	//ラウンドロビンスケジューリングを開始。 sizeが0になるまで続ける。
	while(size > 0){
		
		//display(queue, front, rear);
		
		//先頭のプロセスの残りtimeがこのクオンタム内で終了する::
		if(queue[front].time <= q){
			
			//doneTimeを先頭プロセスの残りtimeだけ進める
			doneTime += queue[front].time;
			
			//終了したプロセスの名前と終了時刻（doneTime）を出力し、このプロセスを削除
			printf("%s %d\n", queue[front].name, doneTime);
			
			dequeue(queue, front, n);
			
			size--;
			
		}
		
		//先頭のプロセスの残りtimeがこのクオンタム内で終了しない::
		else{
			
			//doneTimeをクオンタムだけ進める
			doneTime += q;
			
			//先頭プロセスの残りtimeをクオンタム分減らす
			queue[front].time -= q;
			
			//先頭プロセスはキューの末尾へと移動する
			temp = dequeue(queue, front, n);
			
			enqueue(queue, temp, n);
			
		}
		
	}
	
	free(queue);
	
	return 0;
	
}



void enqueue(ALDS queue[], ALDS data, int n){
	
	queue[rear] = data;
	rear++;
	
	if(rear >= 2*n){
		
		rear = 0;
		
	}
	
	return;
	
}



ALDS dequeue(ALDS queue[], int idx, int n){
	
	ALDS ret = queue[idx];
	front++;
	
	if(front >= 2*n){
		
		front = 0;
		
	}
	
	return ret;
	
}


void display(ALDS data[], int from, int to){
	
	int i;
	
	printf("front::%d, rear::%d\n", from, to);
	
	printf("-----------------------------------------\n");
	
	for(i=from; i<to; i++){
		
		printf("%s %d\n", data[i].name, data[i].time);
		
	}
	
	printf("-----------------------------------------\n");
	
}