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: Time Limit Exceeded)

#include <stdio.h>
int main(){
  long n,m;
  scanf("%ld%ld",&n,&m);
  long a[n],b[n];
  for (long i=0;i<n;i++){
    scanf("%ld%ld",&a[i],&b[i]);
  }
  for (long i=n-1;i>0;i--){
    for (long j=0;j<i;j++){
      if (a[j]>a[j+1]){
        long store;
        store=a[j];
        a[j]=a[j+1];
        a[j+1]=store;
        store=b[j];
        b[j]=b[j+1];
        b[j+1]=store;
      }
    }
  }
  long mon=0;
  for (long i=0;i<n;i++){
    if (m>b[i]){
      mon+=a[i]*b[i];
      m-=b[i];
    }
    else{
      mon+=a[i]*m;
      break;
    }
  }
  printf("%ld\n",mon);
  return 0;
}