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++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
int main(void){
  int n,m,i,j,min,y;
  unsigned long long ans=0;
  cin >> n >> m;
  vector<int> a(n+5,0),b(n+5,0);
  for (i=0;i<n;i++) cin >> a.at(i) >> b.at(i);
  for (i=0;i<n-1;i++){
    min=1000000000;
    for (j=i;j<n;j++) if (min>a.at(j)){
      min=a.at(j);
      y=j;
    }
    swap(a.at(i),a.at(y));
    swap(b.at(i),b.at(y));
  }
  i=0;
  while (1){
    if (m>b.at(i)){
      ans+=(long long)a.at(i)*b.at(i);
      m-=b.at(i);
      i++;
    }else{
      ans+=(long long)m*a.at(i);
      break;
    }
  }
  cout << ans << endl;
  return 0;
}