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 <algorithm>
#include <array>
#include <bitset>
#include <cassert>
#include <cfloat>
#include <chrono>
#include <climits>
#include <cmath>
#include <cstdio>
#include <deque>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <sstream>
#include <stack>
#include <stdexcept>
#include <unordered_map>
#include <unordered_set>
#include <vector>
using namespace std;

template <class T1, class T2>
ostream& operator << (ostream& out, const pair <T1, T2> p)
{
    out << '(' << p.first << ',' << p.second << ')';
    return out;
}

template <class T1, class T2>
istream& operator >> (istream& in, pair<T1, T2> &p)
{
    in >> p.first >> p.second;
    return in;
}

template <class T>
istream& operator >> (istream& in, vector<T> &v)
{
    for (T &x : v)
        in >> x;
    return in;
}

template <class T>
ostream& operator << (ostream& out, const vector<vector<T>> &v)
{
    for (const vector<T> &x : v)
        out << x << '\n';
    return out;
}

template <class T>
ostream& operator << (ostream& out, const vector<T> &v)
{
    for (const T &x : v)
        out << x << ' ';
    return out;
}

long long gcd (long long a, long long b)
{
    if (b > a)
        swap(a, b);
    return (b ? gcd(b, a % b) : a);
}

using ll   = long long;
using pii  = pair<int, int>;
using pll  = pair<long long, long long>;
using tiii = pair<pair<int, int>, int>;
using vi   = vector<int>;
using vl   = vector<long long>;
using vvi  = vector<vector<int>>;
using vvl  = vector<vector<long long>>;

#define F          first
#define S          second
#define First      first.first
#define Second     first.second
#define Third      second
#define mp         make_pair
#define rep(i,a,b) for (int i = (a); i < (b); i++)
#define per(i,b,a) for (int i = (b); i > (a); i--)
#define all(x)     x.begin(), x.end()
#define ret(x)     return cout << x, 0;
#define throwex    throw runtime_error ("Found the error.");

const int h = 1000000007;

signed main()
{
    ios::sync_with_stdio(false);
    #ifdef ONLINE_JUDGE
    cin.tie(nullptr);
    cout.tie(nullptr);
    cerr.setstate(ios::failbit);
    #endif

    string t;
    cin >> t;
    string s;
    ll ans = 0;
    int len = t.size();
    for(char c : t)
    {
        if(c == '1' and !s.empty() and s.back() == '1')
        {
            s.pop_back();
            ans += len;
            len -= 2;
        }
        else
            s.push_back(c);
    }

    int z = 0, one_cnt = 0;
    rep(i,0,s.size())
    {
        if(s[i] == '0')
            z++;
        if(s[i] == '1')
        {
            if(one_cnt)
                z--;
            ans += len - 2 * one_cnt - (z + 1) / 2;
            one_cnt++;
        }
    }
    cout << ans;
}
