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<algorithm>
using lint=int64_t;
using namespace std;

int main()
{
	lint N,M;
	pair<lint,lint> p[100010];
	
	cin >> N >> M;
	for(int i=0;i<N;i++)
		cin >> p[i].first >> p[i].second;
	
	sort(p,p+N);
	
	lint ans=0;
	lint cnt=0;
	for(int i=0;i<N;i++)
	{
		lint tmp=min(p[i].second,M-cnt);
		if(cnt<M)
		{
			ans+=p[i].first*tmp;
			cnt+=tmp;
		}
		if(cnt>=M)
			break;
	}
	
	cout << ans << endl;
	return 0;
}