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,l;
  int count;
  long cost=0;
  int tmp_a,tmp_b;
  int a[100000],b[100000];
  
  scanf("%d%d",&n,&m);
 
  for(int i=0;i<n;i++)
    scanf("%d%d",&a[i],&b[i]);
  
 for (int j=0; j<n; ++j) {
    for (int k=j+1; k<n; ++k) {
      if (a[j] > a[k]) {
        tmp_a =  a[j];
        a[j] = a[k];
        a[k] = tmp_a;
        tmp_b =  b[j];
        b[j] = b[k];
        b[k] = tmp_b;
      }
    }
  }

  
  count=m;
  
  for(l=0;l<n;l++){
    
   if(b[l]>count){
     cost+=a[l]*count;
     count=0;
     break;
   }
   else{
    cost+=a[l]*b[l];
    count-=b[l];
   }
  }
  
  printf("%ld",cost);
  
  
 return 0; 
}