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>
using namespace std;
#include<bits/stdc++.h>
#define lli long long int
#include<vector>
int main(){
  lli n,m,sum=0,i,cost,num,u;
  cin>>n>>m;
  vector< pair<lli,lli> >p(n);
  for(i=0;i<n;i++){
    cin>>cost;
    cin>>num;
    p[i]=make_pair(cost,num);
  }
  sort(p.begin(),p.end());
  lli z=m;
  for(i=0;i<n;i++){
    u=min(z,p[i].second);
    sum=sum+u*p[i].first;
    z=z-u;
  }
  cout<<sum;
  return 0;
}