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

class Main {
	public static void main(String[] $) {
		Scanner sc = new Scanner(System.in);
		int n=sc.nextInt(),m=sc.nextInt();
		long count=0,sum=0,res=0;
		Map<Long, Integer> map = new TreeMap<Long,Integer>();
		for(int i=0;i<n;i++) {
			long key=sc.nextLong();
			int v=sc.nextInt();
			if(map.get(key)==null) {
				map.put(key,v);
			}else {
				map.put(key, map.get(key)+v);
			}
		}
		for(long buff : map.keySet()) {
			count+=map.get(buff);
			if(count<m) {
				sum+=map.get(buff)*buff;
			}else {
				sum+=(map.get(buff)-(count-m))*buff;
				break;
			}
			//System.out.println(buff+" "+map.get(buff));
		}
		System.out.println(sum);
	}
}

