Problem:
Score : 300 points

Problem Statement

There is an empty array.
The following N operations will be performed to insert integers into the array.
In the i-th operation (1≤i≤N), b_i copies of an integer a_i are inserted into the array.
Find the K-th smallest integer in the array after the N operations.
For example, the 4-th smallest integer in the array \{1,2,2,3,3,3\} is 3.

Constraints

1≤N≤10^5

1≤a_i,b_i≤10^5

1≤K≤b_1…+…b_n

All input values are integers.

Input

Input is given from Standard Input in the following format:

N K
a_1 b_1
:
a_N b_N

Output

Print the K-th smallest integer in the array after the N operations.

Sample Input 1

3 4
1 1
2 2
3 3

Sample Output 1

3

The resulting array is the same as the one in the problem statement.

Sample Input 2

10 500000
1 100000
1 100000
1 100000
1 100000
1 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000

Sample Output 2

1

Solution in C (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include<stdio.h>

int main(){
  int i,j,n,k;
  scanf("%d %d",&n,&k);
  int kosuu[n];
  int kazu[n];
  int num_sum=0;
  int num_min=100000;
  int tmp;
  int imanankome=0;
  for(i=0;i<n;i++){
    scanf("%d %d",&(kazu[i]),&(kosuu[i]));
  }
  for(i=0;i<n;i++){
    num_sum+=kosuu[i];
  }
  for(i=0;i<n;i++){
    for(j=i;j<n;j++){
      if(kazu[i]>kazu[j]){
	tmp=kazu[i];
	kazu[i]=kazu[j];
	kazu[j]=tmp;
	tmp=kosuu[i];
	kosuu[i]=kosuu[j];
	kosuu[j]=tmp;
      }
    }
  }

  for(i=0;i<n;i++){
    if(k<=imanankome){
      printf("%d\n",kazu[i-1]);
      return 0;
    }
    else{
      imanankome+=kosuu[i];
      //printf("%d:%d\n",kazu[i],imanankome);
    }
  }

  printf("%d\n",kazu[n-1]);
  return 0;
  
}