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>
 
typedef	long long unsigned	ull;
 
ull a[100000] = { 0 };
ull b[100000] = { 0 };
 
void swap(ull *num1, ull *num2)
{
	ull swp = *num1;
	*num1 = *num2;
	*num2 = swp;
}
 
void sort(ull ary[100000], ull n)
{
	ull i, p;
	
	for(i=0; i<n; i++){	//すべての配列を
		for(p=n-1; 0<p; p--){	//先頭から小さい順に並べ替え
			if(ary[p-1] > ary[p]){
				swap(&ary[p-1], &ary[p]);
				swap(&b[p-1], &b[p]);		// 配列bも共にswap
			}
			if(p == i)
				break;
		}
	}
}
 
int main(void)
{
	ull n, k;
	ull i;
	ull cnt = 0;
	
	scanf("%llu %llu", &n, &k);
	for(i=0; i<n; i++)
		scanf("%llu %llu", &a[i], &b[i]);
	
	sort(a, n);	//値が小さい順に配列を並べ替え
	for(i=0; i<n; i++){
		cnt += b[i];	//配列の数をカウント
		
		if(cnt >= k){
			printf("%llu\n", a[i]);
			break;
		}
	}
	
	return 0;
}