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>

using namespace std;

int main() {

int N,M;
cin>>N>>M;
vector<int> a(N);
vector<int> b(N);
for(int i=0;i<N;i++)
{
   cin>>a[i]>>b[i];
   
}
vector<pair<int,int>> AB(N);
for(int i=0;i<N;i++){
   AB[i]=make_pair(a[i],b[i]);
}
sort(AB.begin(),AB.end());
long long ans=0;
for(int i=0;i<N;i++){
int b=min(M,AB[i].second);
ans+=(long long)b*AB[i].first;
M-=b;
}
cout<<ans<<endl;
   return 0;
}