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<iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <map>
using namespace std;
//typedef pair<long, long> pair;
int main(){
  int N, M;
  cin >> N >> M;

  pair<long,long> p;
  vector<pair<long, int> > AB;

  long a;
  int b;
  for(int i=0;i<N;i++){
    cin >> a >> b;
    p = make_pair(a,b);
    AB.push_back(p);
  }

  sort(AB.begin(), AB.end());

  //cout << AB[0].first <<endl;
  long cost = 0;
  int count = M;
  int i = 0;

  while(count >= AB[i].second){    
    count -= AB[i].second;
    cost += (AB[i].first)*(AB[i].second);
    i++;    
  }

  cost += (AB[i].first)*count;

  cout << cost <<endl;

  return 0;
}