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: Runtime Error)

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

#define REP(i,n) for(i=0;i<n;i++)

int iin() {
    int i;
    scanf("%d", &i);
    return i;
}

long long llin() {
    long long i;
    scanf("%lld", &i);
    return i;
}

double din() {
    double i;
    scanf("%lf", &i);
    return i;
}

char cin() {
    char c;
    scanf("%c", &c);
    return c;
}

void stin(char *s) {
    scanf("%s", s);
}

void iout(int i) {
    printf("%d\n", i);
}

void llout(long long i) {
    printf("%lld\n", i);
}

void dout(double i) {
    printf("%lf\n", i);
}

void cout(char c) {
    printf("%c\n", c);
}

void stout(char *s) {
    printf("%s\n", s);
}

int dfs(int v, int n, int *visited, int table[n][n]) {
    int all_visited = 1, i;

    REP(i,n) {
        if (!visited[i]) {
            all_visited = 0;
            break;
        }
    }

    if (all_visited) {
        return 1;
    }

    int count = 0;

    REP(i,n) {
        if (!table[v][i]) continue;
        if (visited[i]) continue;

        visited[i] = 1;
        count += dfs(i, n, visited, table);
        visited[i] = 0;
    }

    return count;
}

int main() {
    int n = iin(), k = iin(), i, a, b;
    long long c[INT_MAX];
    memset(c, 0, sizeof c);
    REP(i, n) {
        a = iin(), b = iin();
        c[a] += b;
    }
    long long count = 0;
    i = 0;
    while(count < k) {
        i++;
        count += c[i];
    }
    iout(i);
    return 0;
}