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 (Clang 3.8.0) (judge verdict: Wrong Answer)

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

struct stor{
  int a;
  int b;
};

int asc(const void *p, const void *q){
  struct stor *P = (struct stor*)p;
  struct stor *Q = (struct stor*)q;
  return P->a - Q->a;
}

int main(void){
  int n, m;
  scanf("%d %d", &n, &m);
  
  struct stor num[n];
  int i;
  
  for(i = 0; i < n; i++){
    scanf("%d %d", &num[i].a, &num[i].b);
  }
  
  qsort(num, n, sizeof(struct stor), asc);
  
  unsigned long sum = 0, kazu = 0;
  kazu = m;
  for(i = 0; i < n; i++){
    kazu -= num[i].b;
    if(kazu >= 0){
      sum += num[i].a * num[i].b;
      if(kazu == 0){
        break;
      }
    }
    else{
      sum += num[i].a * (num[i].b + kazu);
      break;
    }
  }
  printf("%lu\n", sum);
  return 0;
}