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)

#include <stdio.h>
int main(){
  
  int N, M, i, j, tmp1, tmp2;
  long long Sum;
  Sum = 0;
  scanf("%d", &N);
  scanf("%d", &M);
  
  int A[N], B[N];
  
  for(i = 0; i < N; ++i){
    scanf("%d", &A[i]);
    scanf("%d", &B[i]);
  }
  
  for(i = 0; i < N; ++i){
    for(j = i + 1; j < N; ++j){
      if (A[i] > A[j]) {
        tmp1 =  A[i];
        A[i] = A[j];
        A[j] = tmp1;
        tmp2 = B[i];
        B[i] = B[j];
        B[j] = tmp2;
      }
    }
  }
  
  for(i = 0; i < N; ++i){
    if(B[i] <= M){
      Sum += A[i] * B[i];
      M -= B[i];
    }
    else if(B[i] > M){
      Sum += A[i] * M;
      break;
    }
  }
  
  printf("%lld\n", Sum);
  
  return 0;
}