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();
		Integer[] A = new Integer[N];
		HashMap<Integer,Integer> map = new HashMap<Integer,Integer>();
		for(int i=0;i<N;i++) {
			A[i] = sc.nextInt();
			if (map.containsKey(A[i])) map.put(A[i],map.get(A[i])+sc.nextInt());
			else map.put(A[i], sc.nextInt());
		}
		Arrays.sort(A);
		long ret = 0;
		int sum = 0;
		int count = 0;
		while(sum<M) {
			int a = A[count];
			int b = map.get(a);
			if(sum + b >= M) {
				ret+= (long)(M-sum)*a;
				break;
			}
			else {
				ret += (long)a*b;
				sum += b;
				count++;
				map.put(a,0);
			}
		}

		System.out.println(ret);
	}
}