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<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<vector>
#include<cmath>
#include<algorithm>
#include<map>
#include<queue>
#include<deque>
#include<iomanip>
#include<tuple>
#include<cassert>
using namespace std;
typedef long long int LL;
typedef pair<int,int> P;
typedef pair<LL,int> LP;
const int INF=1<<30;
const LL MAX=1e9+7;

void array_show(int *array,int array_n,char middle=' '){
	for(int i=0;i<array_n;i++)printf("%d%c",array[i],(i!=array_n-1?middle:'\n'));
}
void array_show(LL *array,int array_n,char middle=' '){
	for(int i=0;i<array_n;i++)printf("%lld%c",array[i],(i!=array_n-1?middle:'\n'));
}
void array_show(vector<int> &vec_s,int vec_n=-1,char middle=' '){
	if(vec_n==-1)vec_n=vec_s.size();
	for(int i=0;i<vec_n;i++)printf("%d%c",vec_s[i],(i!=vec_n-1?middle:'\n'));
}
void array_show(vector<LL> &vec_s,int vec_n=-1,char middle=' '){
	if(vec_n==-1)vec_n=vec_s.size();
	for(int i=0;i<vec_n;i++)printf("%lld%c",vec_s[i],(i!=vec_n-1?middle:'\n'));
}

int main(){
	int n,m;
	int i,j,k;
	LL a,b,c;
	string sa;
	vector<int> v1;
	LL s=0;
	cin>>sa;
	n=sa.size();
	for(i=0,j=0;i<n;i++){
		if(j%2==0 && sa[i]=='0'){
			v1.push_back(i);
		}else{
			s+=j/2+1;
			j++;
		}
	}
	i=n;
	reverse(v1.begin(),v1.end());
	v1.push_back(-1);
	for(j=0;j<v1.size();j++){
		a=0,b=0;
		for(i--;i>v1[j];i--){
			c=i-(int)v1.size()-j-1;
			if(sa[i]=='1'){
				if(c%2==0)a++;
				else b++;
			}
		}
		c=(int)v1.size()-j-1;
		a*=c/2+j,b*=(c+1)/2;
		s+=a+b;
	}
	cout<<s<<endl;
}