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<bits/stdc++.h>
using namespace std;
long long N=0,M=0,ans=0,sum=0;


int main()
{
  cin>>N>>M;
  vector<pair<long,long>>A(N);
  for(int i=0;i<N;i++){
    cin>>A[i].first>>A[i].second;
  }
  sort(A.begin(),A.end());
  long long i=0;
  while (ans<M) {
    if(A[i].second==0){
      i++;
    }else{
      A[i].second--;
      sum+=A[i].first;
      ans++;
    }
    //ans++;
  }
  cout<<sum;
  return 0;
}