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<string.h>
#include<stdlib.h>
struct drink {
	long long int a;
	int b
};

int a(const void *c, const void *d) {
	struct drink *x = (struct drink *)c;
	struct drink *y = (struct drink *)d;

	if (x->a > y->a)return 1;
	if (x->a < y->a)return -1;
	return 0;
}

int main() {
	long long int sum = 0;
	int n, m, i;
	struct  drink DATA[100000];

	scanf("%d %d", &n, &m);

	for (i = 0; i < n; i++) 
		scanf("%lld %d", &DATA[i].a, &DATA[i].b);

	qsort(DATA, n, sizeof(struct drink), a);

	for (i = 0; 0 < m; i++) {
		if (DATA->b <= m) {
			m -= DATA[i].b;
			sum += DATA[i].a * DATA[i].b;
		}
		else {
			sum += m * DATA[i].a;
			m = 0;
		}
	}

	printf("%lld\n", sum);
	return 0;
}