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;
string t,s;
vector<int> v;
long long ans;
int main(){
	cin>>t;
	int cnt=count(t.begin(),t.end(),'1');
	for(int i=0;i<=cnt;++i){
		ans+=(i+1)/2;
	}
	for(int i=0;i<t.size();++i){
		s+=t[i];
		int sz=s.size();
		if(sz>=2&&s[sz-1]=='1'&&s[sz-2]=='1'){
			s.pop_back();
			s.pop_back();
			ans+=t.size()-cnt;
		}
	}
	t=s;
	cnt=count(t.begin(),t.end(),'1');
	int cnter=0;
	for(int i=0;i<t.size();++i){
		if(t[i]=='1'){
			ans+=t.size()-cnt-((i+1)/2-(cnter+1)/2);
			cnter++;
		}
	}
	cout<<ans<<'\n';
	return 0;
} 