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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>
#include<list>
#include <algorithm>
#include <string>
#include <math.h>
#include<map>
#include <utility>

#include<queue>
using namespace std;
#define REP(i,k,n) for(int i=k;i<n;i++)
 
#define INF (1e10)
typedef long long unsigned int ll;
#define SIZE_OF_ARRAY(array) (sizeof(array)/sizeof(array[0]))
 

template<typename T>
int vector_finder(std::vector<T> vec, T number) {
  auto itr = std::find(vec.begin(), vec.end(), number);
  size_t index = std::distance( vec.begin(), itr );
  if (index != vec.size()) { // 発見できたとき
    return 1;
  }
  else { // 発見できなかったとき
    return 0;
  }
}

template<typename T>
T gcd(T a,T b){
	if(b>0){
		return gcd(b, a % b);
	}else{
		return a;
	}
}

template<typename T>
T lcm(T a,T b){
	T g = gcd(a, b);
	return a * b / g;
}

void clear( std::queue<pair<int,int> > &q )
{
   std::queue<pair<int,int> > empty;
   std::swap( q, empty );
}

int N, M;
ll ans = INF;
ll K;
int main()
{
	cin >> N >> K;

	ll now = 0;
	ll a[N], b[N];
	REP(i,0,N){
		
		cin >> a[i] >> b[i];
		
	}
	REP(i,0,N){
		now += b[i];
		if(K<=now){
			cout << a[i] << endl;
			return 0;
		}
	}



	return 0;
}