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.Scanner;
import java.util.TreeMap;

public class Main{

	public static void main (String[] args) {

		Main main = new Main();
		main.solve();
	}

	private void solve() {

		Scanner scanner = new Scanner (System.in);

		int N = scanner.nextInt();
		int M = scanner.nextInt();

		TreeMap <Long,Long> costAmount = new TreeMap <Long,Long>();

		for (int index=0; index<N; index++) {
			Long cost = scanner.nextLong();
			Long amount = scanner.nextLong();

			if (costAmount.containsKey(cost)) {
				costAmount.put(cost, costAmount.get(cost)+amount);
			} else {
				costAmount.put(cost, amount);
			}
		}

		Long nowAmount = 0l;
		Long money = 0l;

		for (Long nKey : costAmount.keySet()) {

			Long amount = costAmount.get(nKey);
			Long cost = nKey;

			if (M-nowAmount>amount) {
				nowAmount+=amount;
				money+=(amount*cost);
			} else {
				money+=(M-nowAmount)*cost;
				break;
			}

		}

		System.out.println(money);
	}
}