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
/**
 * code generated by JHelper
 * More info: https://github.com/AlexeyDmitriev/JHelper
 * @author aajisaka
 */

#include<bits/stdc++.h>

using namespace std;

void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 42
#endif

#define SPEED ios_base::sync_with_stdio(false);cin.tie(nullptr)
#define rep(i,n) for(int i=0; i<(int)(n); i++)
#define all(v) v.begin(), v.end()
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

using ll = long long;
using ull = unsigned long long;
using P = pair<ll, ll>;

constexpr long double PI = 3.14159265358979323846264338327950288L;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
// Original: https://github.com/beet-aizu/library

/* range add, range max query
auto f = [](ll a, ll b) { return max(a, b);};
auto g = [](ll a, ll b) { return a+b;};
auto ID = 0LL;
LazySegmentTree<ll, ll> seg(f, g, g, 0, 0);
const int sz = 1<<20;
seg.build(vector<ll>(sz, 0));
*/

template <typename T,typename E>
struct LazySegmentTree{
    using F = function<T(T,T)>;
    using G = function<T(T,E)>;
    using H = function<E(E,E)>;
    int n,height;
    F f;
    G g;
    H h;
    T ti;
    E ei;
    vector<T> dat;
    vector<E> laz;
    LazySegmentTree(F f,G g,H h,T ti,E ei):
        f(f),g(g),h(h),ti(ti),ei(ei){}

    void init(int n_){
      n=1;height=0;
      while(n<n_) n<<=1,height++;
      dat.assign(2*n,ti);
      laz.assign(2*n,ei);
    }

    void build(const vector<T> &v){
      int n_=v.size();
      init(n_);
      for(int i=0;i<n_;i++) dat[n+i]=v[i];
      for(int i=n-1;i;i--)
        dat[i]=f(dat[(i<<1)|0],dat[(i<<1)|1]);
    }

    inline T reflect(int k){
      return laz[k]==ei?dat[k]:g(dat[k],laz[k]);
    }

    inline void propagate(int k){
      if(laz[k]==ei) return;
      laz[(k<<1)|0]=h(laz[(k<<1)|0],laz[k]);
      laz[(k<<1)|1]=h(laz[(k<<1)|1],laz[k]);
      dat[k]=reflect(k);
      laz[k]=ei;
    }

    inline void thrust(int k){
      for(int i=height;i;i--) propagate(k>>i);
    }

    inline void recalc(int k){
      while(k>>=1)
        dat[k]=f(reflect((k<<1)|0),reflect((k<<1)|1));
    }

    void update(int a,int b,E x){
      if(a>=b) return;
      thrust(a+=n);
      thrust(b+=n-1);
      for(int l=a,r=b+1;l<r;l>>=1,r>>=1){
        if(l&1) laz[l]=h(laz[l],x),l++;
        if(r&1) --r,laz[r]=h(laz[r],x);
      }
      recalc(a);
      recalc(b);
    }

    void set_val(int a,T x){
      thrust(a+=n);
      dat[a]=x;laz[a]=ei;
      recalc(a);
    }

    T query(int a,int b){
      if(a>=b) return ti;
      thrust(a+=n);
      thrust(b+=n-1);
      T vl=ti,vr=ti;
      for(int l=a,r=b+1;l<r;l>>=1,r>>=1) {
        if(l&1) vl=f(vl,reflect(l++));
        if(r&1) vr=f(reflect(--r),vr);
      }
      return f(vl,vr);
    }

    template<typename C>
    int find(int st,C &check,T &acc,int k,int l,int r){
      if(l+1==r){
        acc=f(acc,reflect(k));
        return check(acc)?k-n:-1;
      }
      propagate(k);
      int m=(l+r)>>1;
      if(m<=st) return find(st,check,acc,(k<<1)|1,m,r);
      if(st<=l&&!check(f(acc,dat[k]))){
        acc=f(acc,dat[k]);
        return -1;
      }
      int vl=find(st,check,acc,(k<<1)|0,l,m);
      if(~vl) return vl;
      return find(st,check,acc,(k<<1)|1,m,r);
    }

    template<typename C>
    int find(int st,C &check){
      T acc=ti;
      return find(st,check,acc,1,0,n);
    }
};


class EBinaryProgramming {
public:
    void solve(istream& cin, ostream& cout) {
      SPEED;

      string t; cin >> t;
      int n = t.size();
      vector<int> zero(1);
      vector<int> one;
      ll dup = 0;
      ll sumzero = 0;
      rep(i, n) {
        if (t[i] == '0') {
          zero.back()++;
          sumzero++;
        } else {
          if (zero.size() == 1 || zero.back() != 0) {
            zero.push_back(0);
            one.push_back(i);
          } else {
            zero.pop_back();
            one.pop_back();
            dup++;
          }
        }
      }
      ll x = dup*(sumzero+1);

      ll L = 0;
      int zs = zero.size();
      debug(zs);
      rep(i, zs-1) {
        L += zero[i];
        ll R = sumzero - L;

        if (one[i]%2 == 0) {
          x += (L+2)/2;
        } else {
          x += (L+1)/2;
        }
        x += R;
      }
      debug(x);

      ll onecnt = n - sumzero - 1;
      ll d = (onecnt+1)/2;
      x += d*(d+1);
      if (onecnt%2==1) {
        x -= d;
      }
      debug(x);
      cout << x << endl;

    }
};

signed main() {
  EBinaryProgramming solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}