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 LuaJIT (2.0.4) (judge verdict: Wrong Answer)

n,m=io.read("*n","*n","*l")
a={}
for i=1,n do
    a[i]={}
    a[i][1],a[i][2]=io.read("*n","*n")
end
table.sort(a,function(x,y)
    if x[1]~=y[1] then
        return x[1]<y[1]
    else
        return x[2]>y[2]
    end
end)

drink=0
yen=0
for i=1,n do
    if m-drink<a[i][2] then
        yen=yen+a[i][1]*(m-drink)
        break
    else
        drink=drink+a[i][2]
        yen=yen+a[i][1]*a[i][2]
    end
end
print(yen)