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(){
	long long int n,m,i=0,j=0,cnt=0,tmp;


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

      scanf("%lld", &a[i]);
      scanf("%lld", &b[i]);
  }

  for (i=0; 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;
        tmp =  b[i];
        b[i] = b[j];
        b[j] = tmp;
      }
    }
  }
  int money=0;
  for(i=0;i<n;i++){
    for(j=0;j < b[i];j++){
      money += a[i];
      cnt++;
      if(cnt >= m) {
    	   printf("%lld",money);
         return 0;
      }
    }
  }
	return 0;
}