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: Accepted)

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

struct DrinkInfo{
	long long price; 
	long long cnt;
};

struct DrinkInfo dinfo[100000];

int cmp(const void *a, const void *b)
{
	struct DrinkInfo *a_i = (struct DrinkInfo *)a;
	struct DrinkInfo *b_i = (struct DrinkInfo *)b;

	long long cmp_val = a_i->price - b_i->price; 

    if (cmp_val < 0)  return -1;
    if (cmp_val == 0) return 0;
	if (cmp_val > 0)  return 1;	
}

int main(int argc, char *argv[])
{
    long long n, m;
	long long  i;
    scanf ("%lld %lld",  &n, &m);

	for (i = 0; i < n; i++) {
        scanf("%lld %lld", &dinfo[i].price,  &dinfo[i].cnt);
	}

	qsort(dinfo, n, sizeof(struct DrinkInfo), cmp);

	long long  total_price = 0;
	long long  num = m;
	for (i = 0; i < n; i++) {
		if (dinfo[i].cnt >= num) {
			total_price += (num * dinfo[i].price);
			break;
		}
		else {
			total_price += (dinfo[i].cnt * dinfo[i].price);
			num -= dinfo[i].cnt;
		}
	}
	printf("%lld\n", total_price);
	return 0;
}