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: Wrong Answer)

#include <iostream>
#include <vector>
#include <utility>
#include <algorithm>
int main() {
  int M;
  std::cin >> M;
  std::vector<std::pair<int,int>> AB(M);
  for(auto &ab:AB)
    std::cin >> ab.first >> ab.second;
  std::sort(AB.begin(),AB.end(),
  [](std::pair<int,int> &l, std::pair<int,int> &r){
    return l.first > r.first;
  });
  int sum=0;
  int cnt = 0;
  for(auto &ab:AB){
    if(cnt + ab.second>=M){
      sum += ab.first*(M-cnt);
      std::cout << sum << std::endl;
      break;
    }
    cnt += ab.second;
    sum += ab.first*ab.second;
  }
  return 0;
}