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
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=2e5+5;
int n,a,b,c;
char s[N];
int main()
{
    scanf("%s",s+1);
    n=strlen(s+1);
    for(int i=1;i<=n;i++) if(s[i]=='0') b++;
    ll ans=0;
    for(int i=1;i<=n;i++)
        if(s[i]=='0') a++,b--;
        else
        {
            c++;
            if(s[i+1]=='1'){ans+=a+b+1;i++;c++;continue;}
            if(i&1)
                ans+=a/2+1+b;
            else ans+=(a+1)/2+b;
        }
    while(c)
    {
        c--;ans+=(c+1)/2;
    }
    printf("%lld\n",ans);
}
