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 PYTHON):

Solution:
import sys
from collections import deque
import copy
import math
def get_read_func(fileobject):
    if fileobject == None :
        return raw_input
    else:
        return fileobject.readline

def main():
    if len(sys.argv) > 1:
        f = open(sys.argv[1])
    else:
        f = None
    read_func = get_read_func(f);
    input_raw = read_func().strip().split()
    [N, M] = [int(input_raw[0]), int(input_raw[1])]
    D = []
    for i in range(N):
        input_raw = read_func().strip().split()
        [A, B] = [int(input_raw[0]), int(input_raw[1])]
        D.append((A, B))
    D.sort()

    cost = 0L
    for i in range(N):
        if M <= 0:
            break
        M -= D[i][1]
        buy = D[i][1]
        if M < 0:
            buy += M

        cost += long(buy) * D[i][0]
    print cost

if __name__ == '__main__':
    main()