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;
///////////////////////////////////////////
const long long int INF = 1LL<<60;
const long long int Mod = 1000000007;
using ll = long long int; using ci = const int;
using vi = vector<int>;  using Vi = vector<long long int>;
using P = pair<int, int>;  using PLL = pair<ll, ll>;
using matrix = vector<vector<ll>>;
#define pb(x) push_back(x)
#define mp(x,y) make_pair(x,y)
#define all(x) (x).begin(),(x).end()
#define rp(i,N) for(ll i = 0; i < (ll)N; i++)
#define repi(i,a,b) for(ll i = ll(a); i < ll(b); ++i)
template<class T>bool chmax(T &former, const T &b) { if (former<b) { former=b; return true; } return false; }
template<class T>bool chmin(T &former, const T &b) { if (b<former) { former=b; return true; } return false; }
template<class T>T sqar(T x){ return x*x; }//sqrt(x)は平方根;
#define Sort(v) std::sort(v.begin(), v.end(), std::greater<decltype(v[0])>()) //降順でVをソート
#define p_queue(v) priority_queue<v, vector<v>, greater<v> >
template<class T> inline void princ(T x){cout<<x<<" ";}; 
template<class T> inline void print(T x){cout<<x<<"\n";};
template<class T> inline void Yes(T condition){ if(condition) cout << "Yes" << endl; else cout << "No" << endl; }
template<class T> inline void YES(T condition){ if(condition) cout << "YES" << endl; else cout << "NO" << endl; }
///////////////////////////////////////////////////////////////////////////////////

void solve(){
    string t;
    cin >> t;
    ll n=t.size();
    ll sum=0;
    ll ctr=0;
    rp(i,n){
        if(t[i]=='0'){
            if(i&1){
                if(ctr&1){
                    sum+=(ctr+1)/2;
                    ctr++;
                }
                else sum+=ctr/2;
            }else{
                if(ctr%2==0){
                    sum+=ctr/2+1;
                    ctr++;
                }
                else sum+=(ctr+1)/2;
            }
        }
    }
    if(n&1){
        print((n/2)*(n/2+1)+(n+1)/2-sum);
    }
    else print((n/2)*(n/2+1)-sum);
    return;
}
int main(){
    cin.tie(0);ios::sync_with_stdio(false);
    std::cout<<std::fixed<<std::setprecision(30);
    solve();
    return 0;
}