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.Scanner;
import java.util.stream.IntStream;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int N = sc.nextInt();
        int M = sc.nextInt();
        int[] A = new int[N];
        int[] B = new int[N];
        for (int i = 0; i < N; ++i) {
            A[i] = sc.nextInt();
            B[i] = sc.nextInt();
        }

        System.out.println(solve(A, B, M));

        sc.close();
    }

    static long solve(int[] A, int[] B, int M) {
        int[] sortedIndices = IntStream.range(0, A.length).boxed().sorted((i1, i2) -> Integer.compare(A[i1], A[i2]))
                .mapToInt(x -> x).toArray();

        long result = 0;
        for (int index : sortedIndices) {
            int count = Math.min(B[index], M);
            result += (long) A[index] * count;

            M -= count;
        }

        return result;
    }
}