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>

#define MAX 100000

struct energy_drink {
  long a;
  long b;
};

int main()
{
  int n, m;
  struct energy_drink e[MAX];
  
  int i, j;
  struct energy_drink tmp;
  long price = 0;

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

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

  for (i = n - 1; i > 0; i--) {
    for (j = i; j > 0; j--) {
      if (e[j - 1].a > e[j].a) {
	tmp = e[j - 1];
	e[j - 1] = e[j];
	e[j] = tmp;
      }
    }
  }

  for (i = 0; m > 0; i++) {
    price += e[i].a * (m < e[i].b ? m : e[i].b);
    m -= e[i].b;
  }

  printf("%ld\n", price);

  return 0;
}