Murder in closet happened again in a small restaurant and Conan went there to collect evidence with Kogoro. After they reached the restaurant, they got a list of renters that lived in this restaurant recently. Yet the list was so long that they couldn't get the useful information from it. As an assistant of Conan, you must have been ready to help him to get the information on which rooms the renters have lived in.

There are no more than 20 test cases. The first line of each test case contains two integers n and m, indicating the number of renters and the rooms of the restaurant (0 < n, m <= 100). The i-th line of the next n lines contains two integers t1 and t2, the day when they wanted to check in and to leave (0 < t1 < t2 <= 1000).

Each renter rends exactly one room and their check-in days are distinct. Each time a renter came, the owner would give him/her an available room with minimum room number if there were still empty rooms. Otherwise the renter would leave at once and never come back again. Note that rooms are always returned in morning and rented in afternoon. The input is ended with two zeroes.

For each test case, output n lines of integers. The i-th integer indicates the room number of the i-th renter. The rooms are numbered from 1. If someone didn't live in this restaurant, output 0 in the corresponding line.

2 51 32 44 21 52 33 54 50 0
Sample Output
12122
0
題目大意：
第一行給出兩個數字n，m
n代表房客個數m代表房間數
主人給房客安排房間從1開始到m
下面n行代表n個房客的需求 入住時間和不住的時間
我們需要輸出每一個房客的入住的房間號
如果沒有房間住就輸出0
代碼
#include<stdio.h>#include<string.h>#include<stdlib.h>#include<algorithm>#include<iostream>#include<ctype.h>#include<stack>#include<queue>using namespace std;struct nana{    int i,o,r,n;}a[10003];int cmp1(nana a,nana b){    if(a.i==b.i)        return a.o<b.o;    return a.i<b.i;}int cmp2(nana a,nana b){    return a.n<b.n;}int main(){    int n,m;    int r[1003];    while(scanf("%d%d",&n,&m)==2&&n)    {        for(int i=0;i<1003;i++)r[i]=0;        for(int i=1;i<=n;i++)        {           scanf("%d%d",&a[i].i,&a[i].o);           a[i].n=i;        }        sort(a+1,a+n+1,cmp1);        for(int i=1;i<=n;i++)        {            int flag=0;            for(int j=1;j<=m;j++)            {                if(a[i].i>=r[j])                {                    a[i].r=j;                    r[j]=a[i].o;                    flag=1;                    break;                }            }            if(!flag)                a[i].r=0;        }        sort(a+1,a+n+1,cmp2);        for(int i=1;i<=n;i++)        {            PRintf("%d/n",a[i].r);        }    }}