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;

long long sum;
int n,m,cnt;
pair <int, int> p[100005];
int main (){
	cin >> n >> m;
	for (int i = 1; i <= n; i++){
		cin >> p[i].first >> p[i].second;
	}
	sort(p + 1,p + n + 1);
	for (int i = 1; i <= n; i++){
		int cnti = 0;
		while (cnt != m){
			if (cnti == p[i].second){
				break;
			}
			else{
				sum += p[i].first;
			}
			cnt++;
			cnti++;
		}
	}
	cout << sum;
	return 0;
}