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)

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

int asc(int aaa, int bbb) {
	return aaa - bbb;
}

int main() {
	int i, j, k, f = 0, n, s, t, m;
	int a, b, c, d, e;
	double x, y, z;
	int aa[1100000];
	int bb[1100000];
	int cc[1100000];

	scanf("%d %d", &n, &m);
	for (i = 0;i < n;i++) {
		scanf("%d %d", &aa[i], &bb[i]);
	}
	qsort(aa, sizeof(aa) / sizeof(int), sizeof(int), asc);
	/*
	for (i = 0;i < n;i++) {
		for (j = 0;j < n - 1;j++) {
			if (aa[j] < aa[j - 1]) {
				e = aa[j];
				s = bb[j];
				aa[j] = aa[j - 1];
				bb[j] = bb[j - 1];
				aa[j - 1] = e;
				bb[j - 1] = s;
			}
		}
	}
	*/

	x = 0;
	y = 0;
	for (i = 0;y < m;i++) {
		if (y + bb[i] > m) {
			x = x + ((m - y)*aa[i]);
			break;
		}
		y = y + bb[i];
		x = x + (bb[i] * aa[i]);
	}




	printf("%lf", x);

	return 0;
}