Problem:
Score : 300 points

Problem Statement

Hearing that energy drinks increase rating in those sites, Takahashi decides to buy up M cans of energy drinks.

There are N stores that sell energy drinks. In the i-th store, he can buy at most B_i cans of energy drinks for A_i yen (the currency of Japan) each.

What is the minimum amount of money with which he can buy M cans of energy drinks?

It is guaranteed that, in the given inputs, a sufficient amount of money can always buy M cans of energy drinks.

Constraints

All values in input are integers.

1 \leq N, M \leq 10^5

1 \leq A_i \leq 10^9

1 \leq B_i \leq 10^5

B_1 + ... + B_N \geq M

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
A_2 B_2
\vdots
A_N B_N

Output

Print the minimum amount of money with which Takahashi can buy M cans of energy drinks.

Sample Input 1

2 5
4 9
2 4

Sample Output 1

12

With 12 yen, we can buy one drink at the first store and four drinks at the second store, for the total of five drinks. However, we cannot buy 5 drinks with 11 yen or less.

Sample Input 2

4 30
6 18
2 5
3 10
7 9

Sample Output 2

130

Sample Input 3

1 100000
1000000000 100000

Sample Output 3

100000000000000

The output may not fit into a 32-bit integer type.

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

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

void q_sort(int numbers[], int left, int right);

int main(void){
    int n, m;
    int *a;
    int *b;
    int count = 0;
    long pay=0;
    int c=0;
    
    int tempa, tempb;
    
    scanf("%d %d", &n, &m);
    

	a = (int *)malloc(sizeof(int) * n);
	b = (int *)malloc(sizeof(int) * n);

	for (int i = 0; i < n; i++) {
		scanf("%d %d", &tempa, &tempb);
		a[i] = tempa;
		b[i] = tempb;
	}
	
	q_sort(a, 0, n-1);
	q_sort(b, 0, n-1);
	
	/*
	for (int i = 0; i < n; i++) {
	    for (int j = 0; j < i; j++) {
        	    if(a[i] < a[j]){
        	        tempa = a[i];
        	        a[i] = a[j];
        	        a[j] = tempa;
        	        
        	        tempb = b[i];
        	        b[i] = b[j];
        	        b[j] = tempb;
        	    }
    	}
	}
	*/

	while(count < m){
	    pay += a[c];
	    count++;
	    b[c]--;
	    if(b[c]==0) c++;
	}
    
    printf("%ld", pay);
    
    free(a);
    free(b);

}

void q_sort(int numbers[], int left, int right)
{
    int pivot, l_hold, r_hold;

    l_hold = left;
    r_hold = right;
    pivot = numbers[left];
    while (left < right)
    {
        while ((numbers[right] >= pivot) && (left < right))
            right--;
        if (left != right)
        {
            numbers[left] = numbers[right];
            left++;
        }
        while ((numbers[left] <= pivot) && (left < right))
            left++;
        if (left != right)
        {
            numbers[right] = numbers[left];
            right--;
        }
    }
    numbers[left] = pivot;
    pivot = left;
    left = l_hold;
    right = r_hold;
    if (left < pivot)
        q_sort(numbers, left, pivot-1);
    if (right > pivot)
        q_sort(numbers, pivot+1, right);
}