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 <stdlib.h>

int main(void)
{
  int num_of_stores;
  int num_of_drinks;

  scanf("%d%d", &num_of_stores, &num_of_drinks);

  struct node {
    long int price[num_of_stores];
    int limit[num_of_stores];
    int last;
  };

  struct node h;
  h.last = 0;

  int i, j;
  long int temp;
  for(i = 0; i < num_of_stores; i++)
    {
      scanf("%d%d", &h.price[h.last], &h.limit[h.last]);
      j = h.last;
      while(j > 0)
        {
          if(h.price[j] < h.price[(j-1)/2]) {
            temp = h.price[j];
            h.price[j] = h.price[(j-1)/2];
            h.price[(j-1)/2] = temp;

            temp = (long int)h.limit[j];
            h.limit[j] = h.limit[(j-1)/2];
            h.limit[(j-1)/2] = (int)temp;

            j = (j-1)/2;
          }
          else {
            break;
          }
        }
      h.last++;
    }

  int budget = 0;
  int num_so_far = 0;
  for(i = 0; i < num_of_stores; i++)
    {
      if(num_so_far + h.limit[i] < num_of_drinks) {
        budget += h.limit[i] * h.price[i];
        num_so_far += h.limit;
      }
      else {
        budget
          += (num_of_drinks - num_so_far) * h.price[i];
        break;
      }
    }
  printf("%d\n", budget);
  return 0;
}