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)

// I/O
#include <iostream>
#include <iomanip>

// data structures
#include <queue>
#include <set>
#include <vector>

// manipulations
#include <algorithm>

#define rep(i,n) for(int i=0;i< (n);++i)
#define REP(i,n) for(int i=1;i<=(n);++i)
#define for_lowercase(c) for(int c='a';c<='z';++c)
#define for_uppercase(c) for(int c='A';c<='Z';++c)
#define ALL(a) (a).begin(),(a).end()
using LLI = long long int;
LLI GCD(LLI a,LLI b){ if(a<b) return GCD(b,a); else if(a%b==0) return b; else return GCD(b,a%b); }
LLI LCM(LLI a,LLI b){ return (a/GCD(a,b))*b; }
using VI  = std::vector<LLI>;
using PI  = std::pair<LLI,LLI>;
using VVI = std::vector<VI>;
using VPI = std::vector<PI>;
template<class T> bool is_odd (T x){ return   x%2 ; }
template<class T> bool is_even(T x){ return !(x%2); }

int main() {
    std::ios::sync_with_stdio(false);
    LLI sum[100005]={};
    LLI N,K;
    std::cin >> N >> K;

    rep(i,N){
        LLI a,b;
        std::cin >> a >> b;
        sum[a] += b;
    }

    REP(i,100000){
        sum[i] += sum[i-1];
    }
    REP(i,100000){
        if( sum[i-1] < K && K <= sum[i]){
            std::cout << i << std::endl;
            break;
        }
    }
    return 0;
}