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: Runtime Error)

#include "stdio.h"
#include "stdlib.h"
int main()
{
	typedef struct {
		long value;
		long num;
		long nextP;
	}IN_INFO;
	IN_INFO* ptr;
	IN_INFO* ptr_top;

	long N, K;
	long a, b,i,j;
	long inputNum = 0;
	long c = 0;	//挿入場所
	long top = 0;//配列の先頭
	scanf("%ld %ld",&N,&K);

	ptr = (IN_INFO*)malloc(sizeof(IN_INFO) * N);
	ptr_top = ptr;
	scanf("%ld %ld", &a, &b);
	ptr[0].value = a;
	ptr[0].num = b;
	ptr[0].nextP = -1;

	//代入
	for (i = 1; i < N; i++) {
		scanf("%ld %ld", &a, &b);
		ptr[i].value = a;
		ptr[i].num = b;
		//NextP(という名の配列添え字)のつなぎ変え
		for (j = 0; j < N; j++) {
			if (ptr[j].value > a) {
				ptr[i].nextP = ptr[j].nextP;
				ptr[j].nextP = i;
				top = j;
				break;
			}
			if (ptr[j].nextP = -1) {
				ptr[j].nextP = i;
				ptr[i].nextP = -1;
				ptr[i].value = a;
				ptr[i].num = b;
				break;
			}
		}
	}
	//結果表示
	for (i = 0; i < N; i++) {
		K = K - ptr[top].num;
		if (K < 1) {
			printf("%ld",ptr[top].value);
			break;
		}
		top = ptr[top].nextP;
	}
	free(ptr_top);
	return 0;

}