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(void)
{
   int n, m, i, f, x[100000], z=0,t;
   long long int a[100000], b[100000], s=0;

   scanf("%d%d",&n,&m);
   for(i=0;i<n;i++){
      scanf("%lld%lld",&a[i],&b[i]);
      x[i]=i;
   }

   for(i=0;i<n-1;i++)
      for(f=i+1;f<n;f++)
         if(a[x[i]]>a[x[f]]){
            t=x[i];
            x[i]=x[f];
            x[f]=t;
         }

   i=0;

   while(z<=m){
      if(z+b[x[i]]<=m)
         s+=a[x[i]]*b[x[i]];
      else
         s+=a[x[i]]*(m-z);
      z+=b[x[i]];
      i++;
   }
   printf("%lld\n",s);
   return 0;
}