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;
#define ll long long
int n,m;
struct node{
	int a,b;
}_[100001];
ll ans;
bool cmp(node _1,node _2){
	return _1.a<_2.a;
}
int main(){
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;i++)
		scanf("%d%d",&_[i].a,&_[i].b);
	sort(_+1,_+n+1,cmp);
	int i;
	for (i=1;m>=_[i].b;i++){
		ans+=(ll)_[i].b*(ll)_[i].a;
		m-=_[i].b;
	}
	ans+=m*(ll)_[i].a;
	printf("%lld\n",ans);
	return 0;
}