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: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int compar(const int *val1, const int *val2);

int main() {
  int n;
  unsigned long long k;
  int a[100000], b[100000], c[100000];
  scanf("%d %lld", &n, &k);
  int i;
  for (i = 0; i < n; i++) {
    scanf("%d %d", &a[i], &b[i]);
    c[i] = a[i];
  }
  i = 0;
  int j = 0;
  qsort(c, n, sizeof(int), (int (*)(const void *, const void *))compar);
  for (; i < n; i++) {
    if (a[i] == c[j]) {
      if (k <= b[i]) {
        printf("%d", a[i]);
        return 0;
      } else {
        k = k - b[i];
        j++;
        if (j == n) {
          printf("%d", c[n-1]);
          return 0;
        }
        if (a[i] != c[j])
          i = 0;
      }
    }
  }
  printf("%d", c[n-1]);
  return 0;
}

int compar(const int *val1, const int *val2) {
  if (*val1 < *val2) {
    return -1;
  } else if (*val1 == *val2) {
    return 0;
  } else {
    return 1;
  }
}