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:
// C - Energy Drink Collector
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define rp(i,n) for(int i=0;i<(n);i++)

int main(){
	int N,M; cin>>N>>M;

	vector<pair<int,int>> AB;
	rp(i,N){
		int a,b; cin>>a>>b;
		AB.emplace_back(a,b);
	}

	sort(AB.begin(), AB.end());

	ll yen = 0;
	rp(i,N){
		int a,b; tie(a,b) = AB[i];
		int c = min(M,b);
		yen += (ll)c*a;
		M -= c;
		if(M<=0)break;
	}
	cout<< yen <<endl;
}