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.*;
import java.math.*;
import java.math.BigInteger;
class ppo implements Comparable<ppo>{
	int a,b;
	public ppo(int a,int b) {
		this.a=a;
		this.b=b;
	}
	public int compareTo(ppo other) {
		return this.a-other.a;
	}
}
public class Main {
	static final long mod = (int) (1e9 + 7);
	static final int inf = (int) (1e9);
	static final int maxn = (int) (1e5 + 10);

	static int n, m,c;
	static String s;
	public static void main(String[] args) {
		Scanner cin = new Scanner(System.in);
		n=cin.nextInt();m=cin.nextInt();
		ppo arr[] =new ppo [maxn];
		for(int i=1;i<=n;i++) {
			arr[i]=new ppo(cin.nextInt(),cin.nextInt());
		}
		Arrays.sort(arr,1,n+1);
		int num=0;
		long ans=0;
		for(int i=1;i<=n;i++) {
			int ret=Math.min(m-num, arr[i].b);
			num+=ret;
			ans+=(long)ret*(long)arr[i].a;
		}
		System.out.println(ans);
	}
}