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>
#include<stdlib.h>

int main(void) {
  int n, m, i, j, min, tmp, need;
  double ans = 0;
  int *a;
  int *b;
  scanf("%d %d", &n, &m);
  need = m;
  a = (int*)malloc(n*sizeof(int));
  b = (int*)malloc(n*sizeof(int));
  for(i=0; i<n; i++) {
    scanf("%d %d", &a[i], &b[i]);
  }
  // aをソート
  for(i=0; i<n-1; i++) {
    min = i;
    for(j=i+1; j<n; j++) {
      if(a[min] > a[j]) {
        min = j;
      }
    }
    tmp = a[i];
    a[i] = a[min];
    a[min] = tmp;
    tmp = b[i];
    b[i] = b[min];
    b[min] = tmp;
  }
  i=0;
  while(need != 0) {
    if(need <= b[i]) {
      ans += (double)need * a[i];
      need = 0;
    }
    else {
      ans += (double)b[i] * a[i];
      need -= b[i];
    }
    i++;
  }
  printf("%.0lf\n", ans);
  return 0;
}