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 (GCC 5.4.1) (judge verdict: Accepted)

#define _CRT_SECURE_NO_WARNINGS

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>

/* define */
#define Yes(X) puts ( (X) ? "Yes" : "No")
#define YES(X) puts ( (X) ? "YES" : "NO")
#define FOR(i, a, b) for ( int i = (a); i < (b) ; i++)
#define REP(i, n) for ( int i = 0; i < (n) ; i++)
#define max(a, b) (((a) > (b)) ? (a) : (b))
#define min(a, b) (((a) < (b)) ? (a) : (b))
#define max3(a, b, c) ((max((a), (b)) > (c)) ? max((a), (b)) : (c))
#define min3(a, b, c) ((min((a), (b)) < (c)) ? min((a), (b)) : (c))

/* const */
const int MOD = 1000000007;

/* alias */
typedef long long ll;
typedef unsigned  u;

/* function */
int cmp(const void *a, const void *b) {
	return *(int*)a - *(int*)b;
}
int cmp2(const void *a, const void *b) {
	return *(int*)b - *(int*)a;
}
int gcd(int a, int b) {
	if (b != 0)return gcd(b, a % b);
	return a;
}
ll Lgcd(ll a, ll b) {
	if (b != 0) return Lgcd(b, a % b);
	return a;
}
int lcm(int a, int b) {
	return a / gcd(a, b) * b;
}
ll Llcm(ll a, ll b) {
	return a / Lgcd(a, b) * b;
}

int calc(int n, int k) {
	int c = 0;
	REP(i, n / k) REP(j, n / k)if (pow(n - i * k, 2) + pow(n - j * k, 2) <= pow(n, 2))c++;
	c *= 4;
	if (n * 2/ k % 2) c += ((n * 2/ k) - 2) * 2 - 1;
	return c;
}

/* main */
int main() {
	ll n, k, a, b, c[100000] = {};
	scanf("%lld%lld", &n, &k);
	REP(i, n){
		scanf("%lld%lld", &a, &b);
		c[a] += b;
	}
	for (int i = 1;; i++) {
		if (c[i] >= k) {
			printf("%lld", i);
			return 0;
		}
		else k -= c[i];
	}
}