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 Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        C solver = new C();
        solver.solve(1, in, out);
        out.close();
    }

    static class C {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            // 入力
            int n = Integer.parseInt(in.next());
            long k = Long.parseLong(in.next());
            long ans = 0;
            int[] seq = new int[100001];
            int a = 0;
            int b = 0;
            long cnt = 0;
            int pos = 0;
            for (int i = 0; i < n; i++) {
                a = Integer.parseInt(in.next());
                b = Integer.parseInt(in.next());
                seq[a] += b;
            }
            for (int i = 0; i < 100000; i++) {
                if (k <= seq[i]) {
                    ans = i;
                    break;
                }
                k -= seq[i];
            }

            // 出力
            out.println(ans);

        }

    }
}