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>
const int N=2e5+5;
char c[N],d[N];
int n,i,x,y,z,m;
long long ans;
int main(){
	scanf("%s",c+1);n=strlen(c+1);
	for(i=1;i<=n;++i)if(c[i]=='1')++x,ans+=x+1>>1;else ++y;
	for(i=1;i<=n;++i)if(c[i]=='1'){
		if(d[m]=='1')--m;else d[++m]='1';
	}else d[++m]='0';
	x=0;
	for(i=1;i<=y;++i)ans+=n-m>>1;
	for(i=1;i<=m;++i)if(d[i]=='1'){
		++x;
		ans+=(x&1?z>>1:z+1>>1)+y-z;
	}else ++z;
	printf("%lld\n",ans);
}