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 <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef vector<VI> VVI;
typedef vector<VL> VVL;
typedef pair<int, int> P;
typedef pair<ll, ll> PL;
#define FOR(i, a, b) for(int i = 0; i < (a); ++i)
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define all(obj) (obj).begin(), (obj).end()
#define rall(obj) (obj).rbegin(), (obj).rend()
#define debug(x) cout << #x << ": " << x << endl
#define out(x) cout << x << endl
//#define int long long int
const int MOD = 1000000007;
const ll LINF = (ll)1e18 - 1;
const int INF = 1e9 - 1;
const double EPS = 0.000000001;
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};

void print(const vector<int> &v) {
  for(auto x : v) {
    cout << x << " ";
  }
  cout << endl;
}

signed main()
{
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n;
  ll k;
  cin >> n >> k;
  vector<P> p;
  rep(i, n) {
    int a, b;
    cin >> a >> b;
    p.emplace_back(a, b);
  }
  sort(all(p));
  int ans;
  ll count = 0;
  rep(i, n) {
    if(count >= k) break;
    ans = p[i].first;
    count += p[i].second;
  }
  out(ans);
  return 0;
}