2934.   

Election Time

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Time Limit: 3.0 Seconds   

Memory Limit: 65536K

Total Runs: 866   

Accepted Runs: 455    

Multiple test files

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The cows are having their first election after overthrowing the tyrannical Farmer John, and Bessie is one of 

N cows (1 ≤ 

N ≤ 50,000) running for President. Before the election actually happens, however, Bessie wants to determine who has the best chance of winning.

The election consists of two rounds. In the first round, the K cows (1 ≤ K ≤ N) cows with the most votes advance to the second round. In the second round, the cow with the most votes becomes President.

Given that cow i expects to get A_i votes (1 ≤ A_i ≤ 1,000,000,000) in the first round and B_i votes (1 ≤ B_i ≤ 1,000,000,000) in the second round (if he or she makes it), determine which cow is expected to win the election. Happily for you, no vote count appears twice in the A_i list; likewise, no vote count appears twice in the B_i list.

Input

* Line 1: Two space-separated integers: 

N and 

K

* Lines 2..N+1: Line i+1 contains two space-separated integers: A_i and B_i

 

Output

* Line 1: The index of the cow that is expected to win the election.

 

Sample Input

5 33 109 25 68 46 5

Sample Output

5

 

Input details

There are 5 cows, 3 of which will advance to the second round. The cows expect to get 3, 9, 5, 8, and 6 votes, respectively, in the first round and 10, 2, 6, 4, and 5 votes, respectively, in the second.

 

Output details

Cows 2, 4, and 5 advance to the second round; cow 5 gets 5 votes in the second round, winning the election.

 

母鸡表示我们造反并独立了，要选个总统。

输入的2列理，第一列是round1的每个鸡的票数。第二列是round2的每个鸡的票数(only if 它晋级了，所以其实不是很科学啊，你都不知道它会不会晋级怎么就有票数了呢？对吧)

那么反正就sort吧。

#include 
     
      #include 
      
       using namespace std;struct goddamncow{           int r1;           int r2;           int index;};int ca(goddamncow a,goddamncow b){    return a.r1
       
        >m>>n;    for(int i=0;i
        
         >fa[i].r1>>fa[i].r2;            fa[i].index=i+1;    }    sort(fa,fa+m,ca);    sort(fa+m-n,fa+m,cb);    cout<
         
          <