Problem:
Score : 300 points

Problem Statement

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.

Sample Input 1

2 5
4 9
2 4

Sample Output 1

12

With 12 yen, we can buy one drink at the first store and four drinks at the second store, for the total of five drinks. However, we cannot buy 5 drinks with 11 yen or less.

Sample Input 2

4 30
6 18
2 5
3 10
7 9

Sample Output 2

130

Sample Input 3

1 100000
1000000000 100000

Sample Output 3

100000000000000

The output may not fit into a 32-bit integer type.

Solution in Python (3.4.3) (judge verdict: Time Limit Exceeded)

N,M=(map(int,input().split()))

C=[input().split() for i in range(N)]

k=0
l=0#何番目の店
m=0#その店で買う数
money=0
X=0#持ち数

while M>X:
    Min=int(C[0][0])
    for i in C:#どの店で買うか
        if Min>=int(i[0]):
            Min=int(i[0])
            l=k
        k+=1

    for x in range(int(C[l][1])):#何個かえるか

        m+=1
        if M==X+m:#max
            money+=m*Min
            print(money)
            break


    money+=m*Min
    X+=m

    C.pop(l)#取り出す
    #リセット
    k=0
    l=0
    m=0#その店で買う数