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.Comparator;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int M = sc.nextInt();
		long[] A = new long[N];
		long[] B = new long[N];
		for(int i = 0; i < N; i++) {
			A[i] = sc.nextLong();
			B[i] = sc.nextLong();
		}
		sc.close();
		Data[] d = new Data[N];
		for(int i = 0; i < N; i++) {
			d[i] = new Data(A[i], B[i]);
		}
		Arrays.sort(d, new MyComp());
		long ans = 0;
		long res = M;
		for(int i = 0; i < N; i++) {
			if(res - d[i].b <= 0) {
				ans += d[i].a * res;
				break;
			}else {
				ans += d[i].b * d[i].a;
				res -= d[i].b;
			}
		}
		System.out.println(ans);

	}
	static class Data{
		long a, b;
		Data(long a, long b){
			this.a = a;
			this.b = b;
		}
	}
	static class MyComp implements Comparator<Data>{
		public int compare(Data d1, Data d2) {
			if(d1.a < d2.a) {
				return -1;
			}else if(d1.a == d2.a) {
				return 0;
			}else {
				return 1;
			}
		}
	}
}