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.Arrays;
import java.util.Map;
import java.util.Scanner;
import java.util.TreeMap;

public class Main {

	public static void main(String args[]) {

		Scanner sc = new Scanner(System.in);

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

		int[] A = new int[N];
		Map<Integer,Integer> map = new TreeMap<Integer, Integer>();
		for(int i=0;i<N;i++) {
			A[i] = sc.nextInt();
			int b = sc.nextInt();

			if(map.containsKey(A[i])) {
				int temp = map.get(A[i]);
				map.remove(A[i]);
				map.put(A[i], temp+b);
				A[i] = Integer.MAX_VALUE;
			}else {
				map.put(A[i], b);
			}
		}
		Arrays.sort(A);

		long yen = 0;
		for(int i=0;i<A.length;i++) {
			if(M > map.get(A[i])) {
				yen += ((long)A[i] * (long)map.get(A[i]));
				M -= map.get(A[i]);
			}else {
				yen += ((long)A[i] * (long)M);
				break;
			}
		}
		System.out.println(yen);
	}
}