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(),m=sc.nextInt();
        long[] a = new long[n];
        long[] b = new long[n];
        Map<Long,Long> map = new HashMap<>();
        for(int i = 0;i<n;i++){
            a[i]=sc.nextInt();
            b[i]=sc.nextInt();
            if(map.containsKey(a[i])){
                map.put(a[i],map.get(a[i])+b[i]);
            }else {
                map.put(a[i],b[i]);
            }
        }
        Arrays.sort(a);
        long ans = 0;
        long count = 0;
        for(int i = 0;i<n;i++){
            if (i!=0 && a[i] == a[i-1]) continue;
            if(count+map.get(a[i])>m){
                ans+=a[i]*(m-count);
                count=m;
            }else {
                ans+=a[i]*map.get(a[i]);
                count+=map.get(a[i]);
            }
            if(count>=m)break;
        }
        System.out.println(ans);
    }
}