Problem: 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.

Examples

Input

2 5
4 9
2 4


Output

12


Input

4 30
6 18
2 5
3 10
7 9


Output

130


Input

1 100000
1000000000 100000


Output

100000000000000

Solution (in CPP):

Solution:
#include <iostream>
#include <utility>
#include <algorithm>
int main(){
  int n,m;
  long long a,b,c=0;
  std::pair<long long,long long> p[100001];
  std::cin>>n>>m;
  for(int i=0;i<n;++i){std::cin>>a>>b;p[i]={a,b};}
  std::sort(p,p+n);
  for(int i=0;i<n;++i){
    if(p[i].second<m){
      c+=p[i].first*p[i].second;
      m-=p[i].second;
    }
    else{
      c+=p[i].first*m;
      break;
    }
  }
  std::cout<<c<<"\n";
}