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>
typedef long long ll;
const int N=100005;
int n,m,a[N],b[N],id[N];ll ans;
bool cmp(const int&x,const int&y){return a[x]<a[y];} 
int main(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;++i)scanf("%d%d",a+i,b+i),id[i]=i;std::sort(id+1,id+1+n,cmp);
	for(int j=1;j<=n;++j){
		int i=id[j];
		if(m>b[i])m-=b[i],ans+=1ll*a[i]*b[i];
		else{
			ans+=1ll*m*a[i];
			break;
		}
	}
	printf("%lld\n",ans);
	return 0;
}