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;

#define rep(i,n) for(int i=0; i<(n); ++i)
#define rrep(i,n) for(int i=1; i<=(n); ++i)
#define drep(i,n) for(int i=(n)-1; i>=0; --i)
#define srep(i,s,t) for (int i = s; i < t; ++i)
#define pb push_back
#define eb emplace_back
#define sz(x) (int)(x).size()
#define fi first
#define se second
#define v(T) vector<T>
#define imposs {cout<<"-1\n";return 0;}
#define LL_MAX LLONG_MAX
#define LL_MIN LLONG_MIN

typedef unsigned int uint;
typedef long long int ll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef pair<int,int> pii;
typedef vector<pii> vpii;
typedef set<int> si;
typedef multiset<int> msi;
template<typename T>string join(const v(T)&v)
{stringstream s;rep(i,sz(v))s<<' '<<v[i];return s.str().substr(1);}
template<typename T>inline ostream& operator<<(ostream&o,const v(T)&v)
{if(sz(v))o<<join(v);return o;}

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);

  string t;
  cin >> t;

  ll ans = 0;

  int s = 0;

  int l = sz(t);

  rep (i, l) {
    if (t[i] == '1') {
      if ((i+s)%2==0) {
        ans += (l - i) + s / 2;
      } else {
        ans += (s + 1) / 2;
      }
    } else { // '0'
      if ((i+s)%2==0) {
        s += 1;
      }
    }
  }

  cout << ans << '\n';
  return 0;
}
