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 9.2.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
int main(void){
  int M,N;
  long long A[100010];
  long long B[100010];
  int i,j,tmp,tmp2;
  long long ans=0;
  scanf("%d %d",&N,&M);
  for(i=1;i<=N;i++){
    scanf("%lld %lld",&A[i],&B[i]);
  }
  for(i=1;i<=N;i++){
    for(j=i+1;j<=N;j++){
     if(A[i]>A[j]){
      tmp=A[i];
      A[i]=A[j];
      A[j]=tmp;
      tmp2=B[i];
      B[i]=B[j];
      B[j]=tmp2;
    }
   }
 }
 for(i=1;i<=N;i++){
   if(B[i]<M){
     ans=ans+A[i]*B[i];
     M=M-B[i];
   }
   else{
     ans=ans+A[i]*M;
     break;
   }
 }
 printf("%lld\n",ans);
 return 0;
}