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<vector>
#include<algorithm>
using namespace std;

using pll=pair<long long,long long>;
int main(){
    int N;
    long long M;
    cin>>N>>M;
    vector<pll>v(N);
    for(int i=0;i<N;i++)cin>>v[i].first>>v[i].second;
    sort(v.begin(),v.end());

    long long res=0;
    for(int i=0;i<N&&M>0;i++){
        res+=v[i].first*min(M,v[i].second);
        M-=min(M,v[i].second);
    }
    cout<<res<<endl;
}