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 (Clang 3.8.0) (judge verdict: Wrong Answer)

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

long int a[100000][2];

void quick_sort(long int a[100000][2], int i, int j);
int partition(long int a[100000][2], int i, int j);

int main(void)
{
	
	int n, m;
	int i;
	long int sum = 0;
	long int money = 0;
	int count_i;
	
	scanf("%d%d", &n, &m);
	
	for(i = 0; i < n; i++) {
		scanf("%ld%ld", &a[i][0], &a[i][1]);
	}
	
	quick_sort(a, 0, n - 1);
	
	for(i = 0; i < n; i++) {
		sum += a[i][1];
		if(sum >= m) {
			count_i = i;
			break;
		}
	}
	
	sum = 0;
	
	for(i = 0; i < count_i; i++) {
		sum += a[i][1];
		money += a[i][0] * a[i][1];
	}
	
	money += a[count_i][0] * (m - sum);
	
	printf("%ld", money);
	
	return 0;
}

void quick_sort(long int a[100000][2], int i, int j)
{
	int k;
	
	if(i == j) {
		return;
	}
	
	if( (k = partition(a, i, j)) == -1) {
		return;
	}
	
	quick_sort(a, i, k - 1);
	quick_sort(a, k, j);
	
	return;
}

int partition(long int a[100000][2], int i, int j)
{
	int l = i, r = j;
	int index;
	long int k;
	long int work;
	
	k = a[l][0];
	
	if(index == j) {
		return -1;
	}
	
	while(l <= r) {
		
		while(a[l][0] < k && l <= j) {
			l++;
		}
		
		while(a[r][0] > k && r >= i) {
			r--;
		}
		
		if(l > r) {
			break;
		}
		work = a[l][0];
		a[l][0] = a[r][0];
		a[r][0] = work;
		
		work = a[l][1];
		a[l][1] = a[r][1];
		a[r][1] = work;
		
		l++;
		r--;
		
	}
	
	return l;
}