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();
    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();
    }
    Main abc122C = new Main();
    System.out.println(abc122C.solve(N, M, A, B));
  }
 
  public long solve(int N, int M, int[] A, int[] B) {
    long ans = 0;
    Store[] Stores = new Store[N];
    for (int i = 0; i < N; i++) {
      Stores[i] = new Store(A[i], B[i]);
    }
    Arrays.sort(Stores);
    for (int i = 0; i < N; i++) {
      int num = M > Stores[i].num ? Stores[i].num : M;
      ans += (long) num * Stores[i].price;
      M -= num;
      if(M == 0) break;
    }
    return ans;
  }
 
  class Store implements Comparable<Store> {
    int price;
    int num;
    public Store(int price, int num){
      this.price = price;
      this.num = num;
    }
    @Override
    public int compareTo(Store other) {
      return this.price - other.price;
    }
  }
}