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<pair <long,long> > Shop(N);

	for(int i = 0; i < N; i++){
		cin >> Shop[i].first >> Shop[i].second;
	}
	
	sort(Shop.begin(),Shop.end());
	
	long sum = 0,num = 0;
	int a = 0,b = 0;
	
	while(num < M){
		num++;
		sum += Shop[a].first;
		b++;
		
		if(b == Shop[a].second){
			a++;
			b = 0;
		}
	}
	cout << sum << endl;
}