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>
typedef long long ll;
using namespace std;

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	ll n,m,sum=0;
	cin>>n>>m;
	vector<pair<ll,ll>>v(n);
	for(int i=0;i<n;i++){
		cin>>v.at(i).first>>v.at(i).second;
	}
	sort(v.begin(),v.end());
	ll i=0;
	while(m>0){
		if(v.at(i).second>0){
			v.at(i).second--;
			sum+=v.at(i).first;
			m--;
		}else{
			i++;
		}
	}
	cout<<sum<<endl;
}