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: Accepted)

#include <string>
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
//typedef pair<long long,long long> P;
//typedef pair<long long,P> P1;
//typedef pair<P,P> P2;
#define pu push
#define pb push_back
#define mp make_pair
#define eps 1e-7
#define INF 1000000000
#define mod 1000000007
#define fi first
#define se second
#define rep(i,x) for(long long i=0;i<x;i++)
#define repn(i,x) for(long long i=1;i<=x;i++)
#define rrep(i,x) for(long long i=x-1;i>=0;i--)
#define rrepn(i,x) for(long long i=x;i>1;i--)
#define SORT(x) sort(x.begin(),x.end())
#define ERASE(x) x.erase(unique(x.begin(),x.end()),x.end())
#define POSL(x,v) (lower_bound(x.begin(),x.end(),v)-x.begin())
#define POSU(x,v) (upper_bound(x.begin(),x.end(),v)-x.begin())
long long n,h[100005];
ll dp[100005];
#define ALL(v)  (v).begin(),(v).end()
//#define MAX(a,b) if(a>b)
#define MAX(a,b) (((a)>(b))?(a):(b))
#define MIN(a,b) (((a)<(b))?(a):(b))

#define DEBUG
#define pr printf
#define re return
#define CASET int ___T, case_n = 1; scanf("%d ", &___T); while (___T-- > 0)


int GCD(int a, int b) { return b ? GCD(b, a%b) : a; }

stack<long long> sta;   
queue<long long> que;
deque<long long> dque;

set <string> s_set;

int vis[200001];

//最大公約数
int gcd(int x, int y) { return y ? gcd(y, x % y) : x; }
//最小公倍数
int lcm(int x, int y) { return x / gcd(x, y) * y; }
//最大公約数 複数個
int gcd(const vector<int> &v) {
	int ret = v[0];
	for (int i = 1; i < v.size(); i++)
		ret = gcd(ret, v[i]);
	return ret;
}
//最小公倍数 複数個
int lcm(const vector<int> &v) {
	int ret = v[0];
	for (int i = 1; i < v.size(); i++)
		ret = lcm(ret, v[i]);
	return ret;
}

struct Student {
  string name;
  int no, point;
};

// a < b ならtrueを返す
bool comp(const Student& a, const Student& b) {
  if (a.name != b.name)
    return a.name < b.name; // 高い順に並べたいので演算子を逆に
  if (a.point != b.point)
    return a.point > b.point; // 同様に比較演算子を反転
  return a.no < b.no;
}

std::string replaceOtherStr(std::string &replacedStr, std::string from, std::string to) {
    const unsigned int pos = replacedStr.find(from);
    const int len = from.length();
 
    if (pos == std::string::npos || from.empty()) {
        return replacedStr;
    }
 
    return replacedStr.replace(pos, len, to);
}


//	素因数分解
void countPN(int n, std::vector<int> &count)//nを素因数に分解し、countを更新
{
	int m = n;
	for(int i=2 ; i*i<=n ; ++i)
	{
		while(m%i == 0)
		{
			m /= i;
			count[i]++;
		}
		if(m==1) break;
	}
	if(m!=1) count[m]++;

}

int main(){
	
	long long A,B,C,D,E,F,G,H,I,J,L,N,M,K,O,P,Q,R,S,T,U,V,W,X,Y,Z;
	long long sum,sum1,sum2,sum3,sum4;
	sum=sum1=sum2=sum3=sum4=0;
	long long flg,flg1,flg2,flg3,cnt,cnt1,cnt2,cnt3,cnt4;
	flg=flg1=flg2=flg3=cnt=cnt1=cnt2=cnt3=cnt4=0;
	long long max;
	long long max1;
	max=max1=0;
	long long min;
	long long min1;
	min=min1=INF;
	long long wk,wk1,wk2,wk3,wk4;
	wk=wk1=wk2=wk3=wk4=0;
    bool f;
	ll ans=0;
	

	//**************************ここからスタート******************************************
	//std::cin >> N;
	//std::cin >> N >> M;
	//std::cin >> N >> M >> K;
	std::cin >> N >> K;
	//std::cin >> A >> B;
	//std::cin >> A >> B >> C;
    
	//std::vector<long long> v(N);
	//std::vector<long long> w(N);
	//std::vector<long long> x(N);
	//std::vector<int> v(100, 0);
	//int a[26]={0};
	//vector<long long> flagment(N, 0);
	
	//vec.push_back(i);
	//accumulate(vec.begin(), vec.end(), 0)
    //v.resize
    
	// 配列入力1
	/*
	for(long long i=0; i<N; i++){
		std::cin >> v[i];
	}
	*/
	

	
	//●●●  ペア関連
	
	vector<pair<long long, long long> > p(N);
	for(long long i=0; i<N; i++){
		long long a, b;
		cin >> a >> b;
		p[i] = make_pair(a, b);
	}
    
	sort(p.begin(), p.end());
	//sort(p.begin(), p.end(),greater<pair<long long,long long> >());
	
	//●●●	通常（ペア用）
	
	for(long long i=0; i<N; i++){
		sum+=p[i].se;
		if(sum<K){
			
		}
		else{
			pr("%lld",p[i].fi);
			re 0;
		}
	}

	return 0;
}