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>

#define repd(i, a, b) for (ll i = (a); i < (b); i++)
#define rep(i, n) repd(i, 0, n)
typedef long long ll;

using namespace std;

template <typename T> void output(T, int);
int gcd(int a, int b);
void prime_map(map<ll, ll> &, ll);

int main() {
    // source
    ll n, k;
    cin >> n >> k;
    ll a[n], b[n];
    rep(i, n) { cin >> a[i] >> b[i]; }
    vector<pair<ll, ll>> v;
    rep(i, n) { v.push_back(make_pair(a[i], b[i])); }
    sort(v.begin(), v.end());
    ll num = 0;
    rep(i, n) {
        num += b[i];
        if (num >= k) {
            cout << a[i] << endl;
            break;
        }
    }
    return 0;
}

template <typename T> void output(T a, int precision) {
    if (precision > 0) {
        cout << setprecision(precision) << a << "\n";
    } else {
        cout << a << "\n";
    }
}

template <typename T> T min(T a, T b) {

    if (a < b) {
        return a;
    } else {
        return b;
    }
}

int gcd(int a, int b) {
    if (b == 0) {
        return a;
    }
    return gcd(b, a % b);
}

void prime_map(map<ll, ll> &m, ll num) {
    ll cache = num;
    repd(i, 2, sqrt(num) + 2) {
        if (num % i == 0) {
            if (m[i] == 0) {
                m[i] = 1;
            } else {
                m[i]++;
            }

            num /= i;
            break;
        }
    }

    if (cache == num) {
        if (m[num] == 0) {
            m[num] = 1;
        } else {
            m[num]++;
        }
        return;
    } else {
        prime_map(m, num);
    }
}