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;

const int maxn=200005;
int n,cnt,tot;
int suf[2];
bool bad[maxn];
char s[maxn];
long long ans;

int main(){
	scanf("%s",s);n=strlen(s);
	for(int i=0;i<n;i++)if(s[i]=='1'||(s[i]=='0'&&(tot&1))){
		bad[i]=true;
		tot++;
		cnt+=(tot&1)&&(s[i]=='1');
		ans+=cnt;
	}
	for(int i=n-1;i>=0;i--){
		suf[i&1]+=(s[i]=='1');
		if(bad[i]){
			cnt-=(tot&1)&&(s[i]=='1');
			tot--;
		}
		else ans+=cnt+suf[(tot+i)&1];
	}
	printf("%lld\n",ans);
	return 0;
}