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
import java.util.*;

public class Main {

    static Scanner sc = new Scanner(System.in);

    public static void main(String[] args) {
        int N = sc.nextInt();
        int M = sc.nextInt();

        TreeMap<Long, Integer> priceToStock = new TreeMap<>();

        for (int i = 0; i < N; i++) {
            long price = sc.nextLong();
            int stock = sc.nextInt();
            if (priceToStock.containsKey(price)) {
                int prevStock = priceToStock.get(price);
                priceToStock.put(price, stock + prevStock);
            } else {
                priceToStock.put(price, stock);
            }
        }
        long ans = 0;
        for (Map.Entry<Long, Integer> entry: priceToStock.entrySet()) {
            long price = entry.getKey();
            int stock = entry.getValue();
            if (M > stock) {
                //終わらない
                M -= stock;
                ans += price * (long) stock;
            } else {
                //終わり
                ans += price * M;
                break;
            }
        }

        System.out.println(ans);
    }
}
