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<cstdio>
#include<cmath>
#include<algorithm>
#include<cstdlib>
#include<cstring>
using namespace std;
inline int read()
{
	int x=0,f=1;char ch=getchar();
	while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0' && ch<='9')x=x*10+ch-'0',ch=getchar();
	return x*f;
}
int stack[20];
inline void write(int x)
{
	if(x<0){putchar('-');x=-x;}
	if(!x){putchar('0');return;}
	int top=0;
	while(x)stack[++top]=x%10,x/=10;
	while(top)putchar(stack[top--]+'0');
}
inline void pr1(int x){write(x),putchar(' ');}
inline void pr2(int x){write(x),puts("");}
char ss[200010];
int main()
{
    //freopen("a.in","r",stdin);
    //freopen("a.out","w",stdout);
    scanf("%s",ss+1);int n=strlen(ss+1);
    long long ans=0;
    int f0=0,f1=0;
    for(int i=1;i<=n;i++)
	{
		if(ss[i]=='1')
		{
			ans+=(i+1)/2;
			if(i&1)f0++;
			else f1++;
		}
		else {int ul=max(f0,f1);ans+=max(f0,f1),f0=f1=ul;}
	}
    printf("%lld\n",ans);
    return 0;
}