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<vector>
#include<algorithm>
typedef long long ll;
using namespace std;

int main(){
	int n,m,k=0;cin>>n>>m;
  	ll ans=0;
  	vector<pair<ll,ll>> v(n);
  	for(int i=0;i<n;i++) cin>>v[i].first>>v[i].second;
  	sort(v.begin(),v.end());
  	while(m){
      	ll t=v[k].second;
    	if(m>=t)
        {m-=t;
      	ans+=t*v[k].first;}
      	else
        {ans+=m*v[k].first;m=0;}
      	k++;
    }
  	cout<<ans;
}