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
def energy_drink_collector(n,m,dic):
    count = 0
    v = 0
    stores = sorted(dic.keys())

    for i in range(n):
        c = stores[i]
        s = dic[c]
        v += s

        if v >= m:
            _s = s - (v - m)
            count += c * _s
            break
        else:
            count += c * s

    return count

if __name__ == '__main__':
    n,m = map(int,raw_input().split())
    a = [0]*n
    b = [0]*n
    dic = {}
    for i in range(n):
        a[i],b[i] = map(int,raw_input().split())
        dic[a[i]] = 0

    for i in range(n):
        dic[a[i]] = dic[a[i]] + b[i]

    print energy_drink_collector(n,m,dic)