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 <iostream>
#include <iomanip>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <cstdio>
#include <utility>
#include <string>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <deque>
#include <numeric>

using namespace std;

typedef uint64_t u64;
typedef int64_t s64;
typedef uint32_t u32;
typedef int32_t s32;
typedef vector<s32> vs32;
typedef vector<u32> vu32;
typedef vector<s64> vs64;
typedef vector<u64> vu64;

const double PI=3.14159265358979323846;

#define MAX(x, y) ((x) < (y) ? (y) : (x))
#define MIN(x, y) ((x) > (y) ? (y) : (x))

#define rep(i, N) for(int i = 0; i < N; ++i)

#define CEIL(x, y) (((x) + (y) - 1) / (y))
#define MOD 1000000007ULL

#define IN(l, r, x) ((l) <= (x) && (x) < (r))


int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);

    s64 n, k;
    cin >> n >> k;

    vs64 num(1e5 + 1, 0);
    rep (i, n)
    {
        int a, b;
        cin >> a >> b;
        num[a] += b;
    }

    rep (i, 1e5)
    {
        num[i + 1] += num[i];
    }

    cout << lower_bound(num.begin(), num.end(), k) - num.begin() << "\n";
    return 0;
}