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 <vector>
#include <set>
#include <algorithm>
#include <math.h>
#include <string.h>
#include <map>
#include <numeric>
using namespace std;

int main()
{
  long long N,M,a,b,ans=0;
  cin>>N>>M;
  vector <pair<long long, long long> > p(N);
  for(int i=0;i<N;i++){
    cin>>a>>b;
    p[i]=make_pair(a,b);
  }
  sort(p.begin(),p.end());
  long long sum=0;
  for(int i=0;i<N;i++){
    if(p[i].second+sum>=M){
      ans+=(M-sum)*p[i].first;
      break;
    }else{
      ans+=p[i].first*p[i].second;
      sum+=p[i].second;
    }
  }
  cout<<ans<<endl;
}