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<cstdio>
#include<algorithm>
using namespace std;
struct f{
	long long int a,b;
}a[1000000];
bool cmp(f e,f d)
{
	return e.a<d.a;
}
int main()
{
	int i;long long int s=0,m,n;
	scanf("%lld%lld",&n,&m);
	for(i=0;i<n;i++)
	scanf("%lld%lld",&a[i].a,&a[i].b);
	sort(a,a+n,cmp);
	for(i=0;i<n;i++)
	{
		if(m>a[i].b)
		{ 
			m-=a[i].b;
			s+=a[i].b*a[i].a;
		}
		else
		{    
			s+=m*a[i].a;
			break;
		}
	}
	printf("%lld\n",s);
	
}