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: Time Limit Exceeded)

#include <stdio.h>
int a[100000] = {}, b[100000] = {};
int main() {
	int n, m, sum = 0, r = 0;
	long long ans = 0;
	scanf("%d%d", &n, &m);
	for (int i = 0; i < n; i++) {
		scanf("%d%d", &a[i], &b[i]);
	}
	int tmp;
	for (int i = 0; i < n; i++) {
		for (int j = n - 1; j > i; j--) {
			if (a[j] < a[j - 1]) {
				tmp = a[j];
				a[j] = a[j - 1];
				a[j - 1] = tmp;
				tmp = b[j];
				b[j] = b[j - 1];
				b[j - 1] = tmp;
			}
		}
	}
	r = m;
	for (int i = 0;; i++) {
		if (sum + b[i] >= m) {
			ans += (long long)a[i] * r;
			break;
		}
		ans += (long long)a[i] * b[i];
		sum += b[i];
		r -= b[i];
	}
	printf("%lld", ans);
	return 0;
}