Takahashi has an empty string S and a variable x whose initial value is 0.

Also, we have a string T consisting of `0` and `1`.

Now, Takahashi will do the operation with the following two steps |T| times.

* Insert a `0` or a `1` at any position of S of his choice.
* Then, increment x by the sum of the digits in the odd positions (first, third, fifth, ...) of S. For example, if S is `01101`, the digits in the odd positions are `0`, `1`, `1` from left to right, so x is incremented by 2.



Print the maximum possible final value of x in a sequence of operations such that S equals T in the end.

Constraints

* 1 \leq |T| \leq 2 \times 10^5
* T consists of `0` and `1`.

Input

Input is given from Standard Input in the following format:


T


Output

Print the maximum possible final value of x in a sequence of operations such that S equals T in the end.

Examples

Input

1101


Output

5


Input

0111101101


Output

26
t = input()
before_is_zero = True
keeped_ones = 0
n0 = 0
ones = [0] * int(1e5+1) # (number of zeros before 1, is_odd)
ones_index = 0
for i, x in enumerate(t):
    if x == '0':
        n0 += 1
        before_is_zero = True
    elif x == '1':
        if before_is_zero:
            ones[ones_index] = (n0, (i+1)%2)
            ones_index += 1
            before_is_zero = False
        else:
            keeped_ones += 1
            ones_index -= 1
            before_is_zero = True

if before_is_zero:
    if len(ones) > 0:
        ones = ones[:-1]

ans = keeped_ones * n0
for i in range(ones_index):
    n0_before, is_odd = ones[i]
    ans += (n0_before + is_odd)//2 + (n0 - n0_before)
ans += (keeped_ones + (ones_index+1)//2) * (keeped_ones + ones_index//2 + 1)
print(ans)