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)

#import <stdio.h>
int aMin(int * deb, int * fin, int * A)
{
  int min = *deb;
  int i;
  for(i = *deb; i < *fin; i++){
    if(*(A + i) < *(A + min)) min = i;
  }
  return min;
}

void inverser(int * a, int * b)
{
  int tmp = * a;
  * a = * b;
  * b = tmp;
}

int main()
{
  int N, M, i;
  scanf("%d%d",&N,&M);
  int A[N], B[N];
  for(i = 0; i < N; i++) scanf("%d%d",A + i, B + i);
  int nbCan = 0, argent = 0, nbBoutiqueVidee = 0, aM = aMin(&nbBoutiqueVidee, &N, A);
  while(nbCan < M){
    argent += A[aM];
    B[aM] -= 1;
    nbCan +=1;
    if(B[aM] == 0){
      inverser(B + nbBoutiqueVidee, B + aM);
      inverser(A + nbBoutiqueVidee, A + aM);
      nbBoutiqueVidee += 1;
      aM = aMin(&nbBoutiqueVidee, &N, A);
    }
  }
  printf("%d",argent);
}