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 {

	static class pair implements Comparable<pair>{
		public final long left;
		public final long right;

		public pair(long left, long right) {
			this.left = left;
			this.right = right;
		}
		
		public int compareTo(pair p) {
			if(left > p.left)return 1;
			else if(left < p.left)return -1;
			else return 0;
		}

	}

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt(), m = sc.nextInt();
		pair a[] = new pair[n];
		for (int i = 0; i < n; ++i) {
			a[i] = new pair(sc.nextLong(), sc.nextLong());
		}
		sc.close();
		Arrays.sort(a);
		long ans = 0, count = 0;
		for (int i = 0; i < n; ++i) {
			long tmp = a[i].right;
			if (count + tmp >= m)
				tmp = m - count;
			count += tmp;
			ans += a[i].left * tmp;
			if (count >= m)
				break;
		}
		System.out.println(ans);

	}

}