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: Runtime Error)

#include<stdio.h>
#define MAX 200

int main(void)
{
  long int N, M, B[MAX], A[MAX], t, s;
  long long int sum = 0;
  long int i, j;
  
  scanf("%ld %ld", &N, &M);
  
  for(i = 0 ; i < N ; i++)
  {
    scanf("%ld %ld", &A[i], &B[i]);
  }
  
  for(i = 0 ; i < N ; i++)
  {
    for(j = i + 1 ; j < N ; j++)
    {
      if(A[i] > A[j])
      {
        t = A[i];
        A[i] = A[j];
        A[j] = t;
        t = B[i];
        B[i] = B[j];
        B[j] = t;
      }
    }
  }
  
  s = B[0];
  t = 0;
  for(i = 0 ; i < N ; i++)
  {
    if(s < M)
    {
      if(i != 0)
      {
        s += B[i];
        t += 1;
      }
    }
  }
  if(B[0] > M)
  {
    sum = A[0] * M;
  }
  
  for(i = 0 ; i < t ; i++)
  {
    sum += A[i] * B[i];
  }
  sum += A[t] * (M - (s - B[t]));
  
  printf("%ld", sum);
  
  return 0;
}