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 <string>
#include <algorithm>
#include <vector>
#include <array>
#include <queue>
#include <deque>
#include <map>
#include <set>
#include <sstream>
#include <stdio.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <list>
#include <numeric>
#include <stack>
#include <iomanip>

//using namespace std;

#define Rep(i,a,b) for(int i = a; i < b; i++)
#define rep(i,b) Rep(i,0,b)
#define allof(a) (a).begin(), (a).end()

typedef long long ll;
const int inf = 1e9 + 7;
const ll infll = 1ll << 60ll;
const ll mod = 1e9 + 7;
// 0~3までは右左下上 4~7までは斜め
constexpr int dx[] = { 1, 0, -1, 0, 1, 1, -1, -1 };
constexpr int dy[] = { 0, -1, 0, 1, 1, -1, -1, 1 };

namespace {  // 名前なし名前空間
	// gcd, lcm は C++17 でstdに標準入りしたので言語のバージョンに注意
	// 最大公約数
	ll gcd(ll a, ll b) {
		if (b == 0) return a;
		return gcd(b, a % b);
	}

	// 最小公倍数
	ll lcm(ll a, ll b) {
		return a / gcd(a, b) * b;
	}
}

int main() {

	int n, k;
	std::cin >> n >> k;
	std::vector <int> a(n), b(n);
	rep(i, n) std::cin >> a[i] >> b[i];

	std::vector<ll> num(1e5 + 1, 0);
	rep(i, n) num[a[i]] += (ll)b[i];

	ll sum = 0;
	Rep(i, 1, 1e5 + 1) {
		sum += num[i];
		if (sum >= (ll)k) {
			std::cout << i << std::endl;
			return 0;
		}
	}

	return 0;
}