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>
#include<string.h>
 
typedef long long int ll;
#define all(x) (x).begin(), (x).end()
 
using namespace std;

int nxt() {
    int x;
    cin >> x;
    return x;
}

ll nxtl(){
    ll x;
    cin>>x;
    return x;
}
 
void SieveOfEratosthenes(int n,vector <int> &primes) 
{ 
    // Create a boolean array "prime[0..n]" and initialize 
    // all entries it as true. A value in prime[i] will 
    // finally be false if i is Not a prime, else true. 
    bool prime[n+1]; 
    memset(prime, true, sizeof(prime)); 
  
    for (int p=2; p*p<=n; p++) 
    { 
        // If prime[p] is not changed, then it is a prime 
        if (prime[p] == true) 
        { 
            // Update all multiples of p greater than or  
            // equal to the square of it 
            // numbers which are multiple of p and are 
            // less than p^2 are already been marked.  
            for (int i=p*p; i<=n; i += p) 
                prime[i] = false; 
        } 
    } 
  
    // Print all prime numbers 
    for (int p=2; p<=n; p++) 
       if (prime[p]) 
          primes.push_back(p);
} 
 
ll max(ll a,ll b)
{
    if(a>b)
        return a;
    return b;
}
 
ll power(ll x, ll y,ll mod) 
{ 
    ll temp; 
    if( y == 0) 
        return 1; 
    temp = power(x, y/2,mod); 
    if (y%2 == 0) 
        return (temp*temp)%mod; 
    else
        return (((x*temp)%mod)*temp)%mod; 
}

ll binomialCoeff(ll n, ll k) 
{ 
    ll C[k+1]; 
    memset(C, 0, sizeof(C)); 
  
    C[0] = 1;  // nC0 is 1 
  
    for (ll i = 1; i <= n; i++) 
    { 
        // Compute next row of pascal triangle using 
        // the previous row 
        for (ll j = min(i, k); j > 0; j--) 
            C[j] = C[j] + C[j-1]; 
    } 
    return C[k]; 
} 

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ll n=nxt(),k=nxtl();
    vector <pair<int,int> > v(n);
    for(int i=0;i<n;i++)
        cin>>v[i].first>>v[i].second;
    sort(all(v));
    ll cur=v[0].second;
    int i=0;
    while(k>cur)
    {
        cur+=v[++i].second;
    }
    cout<<v[i].first<<endl;

    
    
    return 0;   
}