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;

int main(){
  int n,m;
  cin>>n>>m;
  vector<pair<int64_t,int64_t>> p(n);
  for(int i=0;i<n;++i) cin>>p[i].first>>p[i].second;
  sort(p.begin(),p.end());

  int64_t sum=m,ans=0;
  for(int i=0;i<n;++i){
    int64_t a=p[i].first,b=p[i].second;
    if(b<m){
      ans += a*b;
      m -= b;
    }
    else{
      ans += a*m;
      cout<<ans<<endl;
      break;
    }
  }
}