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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

/*
* Author. Silviase(@silviasetitech)
* For AtCoder
*/

import java.util.*;
import java.lang.*;
import java.math.*;


public class Main {
    
    public static void main(String args[]){
        Scanner sc = new Scanner(System.in);


        int n = sc.nextInt();
        long k = sc.nextLong();
        long[] a = new long[200000];
        long[] aruiseki = new long[200000];
        for (int i = 0; i < n; i++) {
            int tmpa = sc.nextInt();
            int tmpb = sc.nextInt();
            a[tmpa] += tmpb;
        }
        for (int i = 0; i < aruiseki.length; i++) {
            if (i > 0) {
                aruiseki[i] = aruiseki[i-1] + a[i];
                if (aruiseki[i] >= k) {
                    System.out.println(i);
                    break;
                }
            }
        }
        sc.close();
    }


    public static long gcd(Long m, long n) {
        if(m < n) {
            return gcd(n, m);
        }
        if(n == 0){
            return m;
        } 
        return gcd(n, m % n);
    }

    public static long[][] Combination_nCk(long n, long k) {
        n++; k++;
        long[][] ans = new long[(int)n][(int)k];
        for (int i = 0; i < n; i++) {
            ans[i][0] = 1;
        }
        for (int i = 0; i < n-1; i++) {
            for (int j = 0; j < k-1; j++) {
                if (i < j) {
                    ans[i][j] = 0;
                }else{
                    ans[i+1][j+1] = ans[i][j] + ans[i][j+1];
                }
            }
        }
        return ans;
    }

    public static double[][] Probability_nCk(int n, int k) {
        n++; k++;
        double[][] ans = new double[n][k];
        for (int i = 0; i < n; i++) {
            if (i == 0) {
                ans[i][0] = 1;
            }else{
                ans[i][0] = ans[i-1][0]/2;
            }
        }
        for (int i = 0; i < n-1; i++) {
            for (int j = 0; j < k-1; j++) {
                if (i < j) {
                    ans[i][j] = 0;
                }else{
                    ans[i+1][j+1] = (ans[i][j] + ans[i][j+1])/2;
                }
            }
        }
        return ans;
    }
}