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: Accepted)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#define MAX(X, Y) (((X) > (Y))? (X) : (Y))
#define MIN(X, Y) (((X) < (Y))? (X) : (Y))
int compare_int(const void *a, const void *b){
  return *(long int*)a - *(long int*)b; // 小さい順
}
int compare_int_small(const void *a, const void *b){
  return *(long int*)a - *(long int*)b; // 小さい順
}
// 公約数
long int gcd(long int a, long int b){
  if(b == 0){
    return a;
  }else{
    return gcd(b, a % b);
  }
}
// 最小公倍数 = a * b / 最大公約数
// 公倍数
long int lcm(long int a, long int b){
  return a * b / gcd(a, b);
}


// min
int main(void){
  long int n, k;
  long int ans = 0;
  long int a[100001];
  long int b[100001];
  long long int cnt[100001] = {0};

  // nとkを取得して
  scanf("%ld%ld", &n, &k);

  // a[], b[] も取得する
  for(long int i = 0; i < n; i++){
    scanf("%ld%ld", &a[i], &b[i]);
    // どの値が何個あるか格納しておく
    cnt[a[i]] += b[i];
  }

  /* 
  for(long int i = 0; i < n; i++){
    for(long int j = 0; j < b[i]; j++){
      cnt[a[i]]++;
    }
  } 
  */
  //  qsort(c, n, sizeof(long long int), compare_int);
  //  qsort(r, n, sizeof(long int), compare_int_small)  //  fprintf(stderr, "l[0] = %ld  r[0] = %ld\n", l[0], r[0]);

  long long int j = 0;
  // kが0以下になった時の値が答えになる
  while(k > 0){
    fprintf(stderr, "j = %lld\n", j);
    ans = j;
    k -= cnt[j];
    j++;
  }
  
  
  printf("%ld\n", ans);
  
  return 0;
}