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: Accepted)

#include <bits/stdc++.h>

using namespace std;
long long N,M;
struct Veb{
  long long A,B;
};

vector<Veb > v;

bool com(Veb v1, Veb v2) {
  if(v1.A < v2.A) return true;
  else return false;
}



int main() {
  long long a,b;
  ios::sync_with_stdio(false);
  cin >> N >> M;
  Veb tmp;
  for(int i = 0; i < N; i++) {
    cin >> a >> b;
    tmp.A = a;
    tmp.B = b;
    v.push_back(tmp);
  }
  sort(begin(v), end(v), com);

  long long s = 0;
  long long honsu = 0;
  int i = 0;
  for(i = 0; i < N; i++){
    s += v[i].A*v[i].B;
    honsu += v[i].B;
    if(honsu >= M){
      s -= v[i].A*v[i].B;
      honsu -= v[i].B;
      break;
    }
  }
  long long diff = M - honsu;
  s += diff*v[i].A;
  cout << s;
}