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;

#define rep(i,n) for(int i = 0; i < n; i++)
#define REP(i,n) for(int i = 1; i < n; i++)
#define per(i,n) for(int i = n; i >= 0; i--)
#define sz(x) int(x.size())
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()
#define YesorNo(a) printf(a ? "Yes\n" : "No\n")
#define endl '\n'
#define fi first
#define se second
using ll = long long;
using P = pair<int,int>;
using Pl = pair<ll,ll>;
using vi = vector<int>;
using vii = vector<vi>;
using vl = vector<ll>;
using vll = vector<vl>;
using vs = vector<string>;
using vb = vector<bool>;
using vc = vector<char>;
using vcc = vector<vc>;
const int dx[] = {0,1,0,-1,1,1,-1,-1};
const int dy[] = {1,0,-1,0,1,-1,-1,1};
const int inf = (1<<30)-1;
const ll infll = (1LL<<62)-1;
ll ceil(const ll &a, const ll &b){return ((a)+(b)-1)/b;}
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }



int main(){
  ios::sync_with_stdio(false);
  cin.tie(nullptr);


  int n, k;
  cin >> n >> k;
  vector<P> vp;
  rep(i,n) {
    int a, b;
    cin >> a >> b;
    vp.emplace_back(a,b);
  }
  sort(all(vp));
  for(P p : vp) {
    if(k <= p.se) {
      cout << p.fi << endl;
      return 0;
    }
    k -= p.se;
  }
}