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 JAVA):

Solution:
import java.util.*;

public class Main{
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        TreeMap<Integer, Integer> map = new TreeMap<>();
        for (int i = 0; i < n; i++) {
            int a = sc.nextInt();
            int b = sc.nextInt();
            int count = map.getOrDefault(a, 0);
            map.put(a, count + b);
        }

        long count = 0;
        for (Map.Entry<Integer, Integer> e : map.entrySet()) {
            int price = e.getKey();
            int num = e.getValue();
            if (m == 0) break;
            int min = Math.min(m, num);
            count += (long) min * price;
            m -= min;
        }
        System.out.println(count);
    }
}