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<vector>
#include<set>
#include<map>
#include<queue>
#include<string>
#include<algorithm>
#include<iostream>
#include<bitset>
#include<functional>
#include<numeric>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cassert>
#include<cmath>
#include<random>
#include<ctime>
#include<complex>
using namespace std;
typedef long long LL;
mt19937 gene(233);
typedef complex<double> Complex;
#define fi first
#define se second
#define ins insert
#define pb push_back
inline char GET_CHAR(){
	const int maxn = 131072;
	static char buf[maxn],*p1=buf,*p2=buf;
	return p1==p2&&(p2=(p1=buf)+fread(buf,1,maxn,stdin),p1==p2)?EOF:*p1++;
}
inline int getInt() {
	int res(0);
	char c = getchar();
	while(c < '0') c = getchar();
	while(c >= '0') {
		res = res * 10 + (c - '0');
		c = getchar();
	}
	return res;
}

inline LL fastpo(LL x, LL n, LL mod) {
	LL res(1);
	while(n) {
		if(n & 1) {
			res = res * (LL)x % mod;
		}
		x = x * (LL) x % mod;
		n /= 2;
	}
	return res;
}

inline string itoa(int x, int width = 0) {
  string res;
  if(x == 0) res.push_back('0');
  while(x) {
    res.push_back('0' + x % 10);
    x /= 10;
  }
  while((int)res.size() < width) res.push_back('0');
  reverse(res.begin(), res.end());
  return res;
}
template<const LL mod> struct MI{
	LL a;
	MI operator + (const MI & b) {
		MI res{a + b.a};
		if(res.a >= mod) res.a -= mod;
		return res;
	}
	MI operator - (const MI & b) {
		MI res{a - b.a};
		if(res.a <= 0) res.a += mod;
		return res;
	}
	MI operator * (const MI & b) {
		return MI{a * b.a % mod};
	}
	MI operator / (const MI & b) {
		return MI{a * fastpo(b.a, mod - 2, mod) % mod};
	}
};
const int N = 300033;
const int LOG = 20;
const int mod = 1e9 + 7;
const int inf = 1e9 + 7;
int n, m;
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
int f[N];
int main() {
	string t;
	cin >> t;
	int n = t.size();
	int sta = 1;
	int cnt = 0;
	LL ans = 0;
	for(int i = 0; i < n; i++) {
		if(sta) {
			if(t[i] == '1') {
				cnt++;
				ans += cnt;
				f[i] = true;
				sta = 0;
			}
		}else {
			ans += cnt;
			f[i] = true;
			sta = 1;
		}
	}
	int len = 0;
	for(int i = 0; i < n; i++) len += f[i];
	int c0 = 0, c1 = 0;
	for(int i = n - 1; i >= 0; i--) {
		if(!f[i]) {
			swap(c0, c1);
			if(t[i] == '1') c0++;
			ans = ans + (len + 1) / 2 + (len % 2 == 0 ? c0 : c1);
		}else {
			len--;
			swap(c0, c1);
			if(t[i] == '1') c0++;
		}
	}
	cout << ans << endl;
}




