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>

int main(){
  int N,M,i,j,temp;
  double B_sum=0,A_sum=0;
  int A[100000];
  int B[100000];
  scanf("%d %d",&N,&M);
  
  
  for(i=0;i<N;i++){
    scanf("%d %d",&A[i],&B[i]);
  }
  
  for(j=0;j<1000000;j++){
    for(i=0;i<N-1;i++){
      if(A[i]>A[i+1]){
      	temp = A[i];
        A[i] = A[i+1];
        A[i+1] = temp;
        
        temp = B[i];
        B[i] = B[i+1];
        B[i+1] = temp;
      }
    }
  }
  
  for(i=0;i<N;i++){
    //printf("A[%d]%d B[%d]%d\n",i,A[i],i,B[i]);
    B_sum += B[i];
    A_sum += A[i]*B[i];
    if(B_sum >= M) break;
  }
  A_sum = A_sum - A[i]*(B_sum-M);
  
  printf("%.0f",A_sum);
  return 0;
}