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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<cstdio>
#include<algorithm>
 
int main(int argc, char const *argv[]) {
  int n,m;
  long ans=0;
  scanf("%d %d",&n,&m);
  int a[n],b[n];
  for(int i = 0 ;i<n;i++){
    scanf("%d %d",&a[i],&b[i]);
  }
  for(int i=0;i<n;i++){
    int min = i;
    for(int j=i+1;j<n;j++){
      if(a[min]>a[j]){
        int box=a[min];
        a[min]=a[j];
        a[j]=box;
        box=b[min];
        b[min]=b[j];
        b[j]=box;
        b[min]=box;
        min=j;
      }
    }
  }
  int buycount=0;
  for(int i=0;i<n;i++){
    if(buycount+b[i]<=m){
      ans+=a[i]*b[i];
      buycount+=b[i];
    }else{
      ans+=(m-buycount)*a[i];
    }
  }
  printf("%ld",ans);
  return 0;
}