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 "bits/stdc++.h"
using namespace std;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using vi = vector<int>;
using vl = vector<ll>;
using vvi = vector<vi>;
using vvl = vector<vl>;
const int INF = 1 << 28;
const ll MOD = 1'000'000'007;
template <class T> bool chmax(T &a, const T &b) { return (a < b) ? (a = b, 1) : 0; }
template <class T> bool chmin(T &a, const T &b) { return (b < a) ? (a = b, 1) : 0; }

int main() {
    int n, k;
    cin >> n >> k;
    vi a(n), b(n);
    for (int i = 0; i < n; ++i) {
        cin >> a[i] >> b[i];
    }
    map<int, ll> m;
    for (int i = 0; i < n; ++i) {
        m[a[i]] += b[i];
    }
    int ret;
    for (auto mm : m) {
        k -= mm.second;
        if (k <= 0) {
            ret = mm.first;
            break;
        }
    }
    cout << ret << endl;
    return 0;
}