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;
#define int long long
const int N=2e5+5;
char s[N];
int a,b,c,ans;
signed main(){
	cin>>(s+1);int n=strlen(s+1);
	for(int i=1;i<=n;++i)
		if(s[i]=='0') ++b;
	for(int i=1;i<=n;++i)
		if(s[i]=='0') ++a,--b;
		else{
			++c;
			if(s[i+1]=='1'){
				ans+=1+a+b;++c;++i;continue;
			}
			if(i&1) ans+=1+(a/2)+b;
			else ans+=(a+1)/2+b; 
		}
	while(c) ans+=(c/2),--c;
	cout<<ans;
	return 0;
} 