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: Wrong Answer)

import java.util.*;
import java.util.Map.Entry;
 
 
 
public class Main{
	
	
	
	public static Set<Long> set=new HashSet<Long>();
	public static long T;
	public static long ans;
	

 
	
	//最大公約数・最小公倍数（セットで使う）
	public static long gcm(long a, long b) {
		long result = a;
		long k = 0;
		long n = b;
		do {
			k = result % n;
			result = n;
			n = k;
		} while(k != 0);
		return result;
	}
	
	public static long lcm(long a, long b) {
		long g;
		g = gcm(a, b);
		return a/g*b;
	}
	
	static long gcd(long a, long b) {
		if (b == 0) {
			return a;			
		}
		return gcd(b, a % b);
	}
	
	//等差数列
	static long tousasum(int n) {
		return (long)n*((long)n+1)/2;
	}
 
 
 
	//素数判定
	static boolean isPrime (int n) {
		if (n==2) return true;
		if (n<2 || n%2==0) return false;
		double d = Math.sqrt(n);
		for (int i=3; i<=d; i+=2) if(n%i==0){return false;}
		return true;
	}
 
 
	//倍数判定（10進数以外のときに有用）
	static boolean isMultiple (String s, int base, int m) {
		int temp = 0;
		for (int i=0; i<s.length(); i++) {
			temp = (temp*base+Character.getNumericValue(s.charAt(i)))%m;
		}
		if (temp==0) {return true;}
		return false;
	}
 
 
	//階乗
	static long factorial (int i) {
		if (i==1) {return 1;}
		else {return i*factorial(i-1);}
	}
 
 
	//進数変換
	static String toNbase (String sm, int m, int n) {
		return Long.toString(Long.parseLong(sm,m),n);
	}
	
	public static final long gcd_n(long[] param) {
	    final int len = param.length;
	    long g = gcd(param[0], param[1]);    //gcd(a, b) は以前作ったもの
	    for (int i = 1; i < len - 1; i++) {
	        g = gcd(g, param[i + 1]);       //gcd(a, b) は以前作ったもの
	    }
	    return g;
	}
	
	public static long lcm_n(long[] numbers) {
		long l;
		l = numbers[0];
		for (int i = 1; i < numbers.length; i++) {
			l = lcm(l, numbers[i]);
		}
		return l;
	}
	
    public static boolean tree(long N,long r) {
    	ans = r;
    	if(N==T) {
    		return true;
    	}else if(2*N==T||2*N+1==T) {
    		ans = r+1;
    		return true;
    	}else if(2*N>T) {
    		return false;
    	}
    	return tree(2*N,r+1)||tree(2*N+1,r+1);
    	
    	
    }
	
	
 public static void main(String[] args) {
	 Scanner sc=new Scanner(System.in);
	 Map<Integer,Integer> map=new TreeMap<Integer,Integer>();
	 int N=sc.nextInt();
	 long K=sc.nextLong();
	 for(int i=0;i<N;i++) {
		 int a=sc.nextInt(),b=sc.nextInt();
		 if(map.containsKey(a)) {
			 map.put(a,map.get(a)+b);
		 }else {
			 map.put(a,b);
		 }
	 }
	 long result=0;
	 for (Integer nKey : map.keySet())
     {
         result+=map.get(nKey);
         if(result>=K) {
        	 System.out.println(nKey);
        	 break;
         }
     }
    
}
}