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>
#define SIZE 200005
using namespace std;
typedef long long int ll;
typedef pair <int,int> P;

char str[SIZE];
int dp[2];

int main()
{
	scanf("%s",&str);
	int n=strlen(str);
	ll ret=0;
	dp[0]=dp[1]=0;
	for(int i=0;i<n;i++)
	{
		if(str[i]=='0')
		{
			ret+=max(dp[0],dp[1]);
			int m=max(dp[0],dp[1]);
			dp[0]=dp[1]=m;
		}
		else if(str[i]=='1')
		{
			ret+=i/2+1;
			int o=max(dp[1],dp[0]+1);
			int z=max(dp[0],dp[1]);
			dp[0]=z,dp[1]=o;
		}
	}
	printf("%lld\n",ret);
}