Rank List

    这是我第二次做codeforces的比赛，成绩没第一次好，主要原因还是那个万年不变的原因——看不懂题意。报告一下时间2012年3、24 1:12。先不唠叨，很有必要感谢我的上铺兄弟zhifa,他的英语水平很ok，就让他帮我看题，我太心急了，在没有了解题意的情况下就敲代码，结果呢？结果悲剧了！就又一次读题，终于明白了题意，再次感谢他！（童鞋们，还是学英语吧！谁让祖国不强大，啥时比赛都有中英对照时，哈哈。又幻想了.....)
       读题注意以下几点：
      如果解题数目和所用时间都相等则他们就共同排名在（y+1,y+2,..y+x),y是比他们成绩好的，x是同名次的人数
       我的思路先排序，再比较，这个方法很笨拙，先贴上，改天看看大神的

A. Rank List

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Another programming contest is over. You got hold of the contest's final results table. The table has the following data. For each team we are shown two numbers: the number of problems and the total penalty time. However, for no team we are shown its final place.

You know the rules of comparing the results of two given teams very well. Let's say that teama solvedp_a problems with total penalty timet_a and teamb solvedp_b problems with total penalty timet_b. Teama gets a higher place than team b in the end, if it either solved more problems on the contest, or solved the same number of problems but in less total time. In other words, teama gets a higher place than team b in the final results' table if eitherp_a > p_b, orp_a = p_b andt_a < t_b.

It is considered that the teams that solve the same number of problems with the same penalty time share all corresponding places. More formally, let's say there is a group ofx teams that solved the same number of problems with the same penalty time. Let's also say thaty teams performed better than the teams from this group. In this case all teams from the group share placesy + 1,y + 2,...,y + x. The teams that performed worse than the teams from this group, get their places in the results table starting from they + x + 1-th place.

Your task is to count what number of teams from the given list shared the k-th place.

<P< p><P< p>

Input

<P< p>

The first line contains two integers n andk (1 ≤ k ≤ n ≤ 50). Thenn lines contain the description of the teams: thei-th line contains two integersp_i andt_i (1 ≤ p_i, t_i ≤ 50) — the number of solved problems and the total penalty time of the i-th team, correspondingly. All numbers in the lines are separated by spaces.

<P< p><P< p>

Output

<P< p>

In the only line print the sought number of teams that got the k-th place in the final results' table.

<P< p><P< p>

Sample test(s)

<P< p><P< p>

Input

7 24 104 104 103 202 12 11 10

Output

3

Input

5 43 13 15 33 13 1

Output

4

<P< p><P< p>

Note

<P< p>

The final results' table for the first sample is:

• 1-3 places — 4 solved problems, the penalty time equals 10
• 4 place — 3 solved problems, the penalty time equals 20
• 5-6 places — 2 solved problems, the penalty time equals 1
• 7 place — 1 solved problem, the penalty time equals 10

The table shows that the second place is shared by the teams that solved 4 problems with penalty time 10. There are 3 such teams.

The final table for the second sample is:

• 1 place — 5 solved problems, the penalty time equals 3
• 2-5 places — 3 solved problems, the penalty time equals 1

The table shows that the fourth place is shared by the teams that solved 3 problems with penalty time 1. There are 4 such teams.

这是pony神的程序，第二天才发现我的没过

#include<stdio.h>struct node{int score;int time;}p[1000];int main(){int n,k;int t;struct node q;scanf("%d%d",&n,&k);int i,j;for(i=0;i<n;i++)scanf("%d%d",&p[i].score,&p[i].time);for(i=0;i<n-1;i++){for(j=i+1;j<n;j++){if(p[i].score<p[j].score){q=p[i];p[i]=p[j];p[j]=q;}else if(p[i].score==p[j].score){if(p[i].time>p[j].time){q=p[i];p[i]=p[j];p[j]=q;}}}}int s=0;for(i=0;i<n;i++)if(p[i].score==p[k-1].score&&p[i].time==p[k-1].time)s++;printf("%d\n",s);return 0;}--------------------------------------------------------------------------------