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;
  scanf("%d",&n);
  scanf("%d",&m);
  int a[n],b[n];
  int aa[n],bb[n];
  int hi=0,min=1000000001;
  for(int i=0;i<n;i++){
    scanf("%d",&a[i]);
    scanf("%d",&b[i]);
    hi=a[i];
    if(hi<min){
      min=hi;
      for(int j=0;j<i;j++){
        aa[j+1]=aa[j];
        bb[j+1]=bb[j];
      }
      aa[0]=a[i];
      bb[0]=b[i];
    }
    else{
      aa[i]=a[i];
      bb[i]=b[i];
    }
  }
  int sum=0,count=0;
  long long int ans=0;
  while(sum<m){
    sum+=bb[count];
    ans+=bb[count]*aa[count];
    count++;
    //printf("%d %lld\n",sum,ans);
  }
  ans-=(sum-m)*aa[count-1];
  printf("%lld\n",ans);
}