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;
using ll = long long;
using uint = unsigned int;
#define rep(i,n) for(int i=0;i<int(n);i++)
#define rep1(i,n) for(int i=1;i<=int(n);i++)
#define per(i,n) for(int i=int(n)-1;i>=0;i--)
#define per1(i,n) for(int i=int(n);i>0;i--)
#define all(c) c.begin(),c.end()
#define si(x) int(x.size())
#define pb emplace_back
#define fs first
#define sc second
template<class T> using V = vector<T>;
template<class T> using VV = vector<vector<T>>;
template<class T,class U> void chmax(T& x, U y){if(x<y) x=y;}
template<class T,class U> void chmin(T& x, U y){if(y<x) x=y;}
template<class T> void mkuni(V<T>& v){sort(all(v));v.erase(unique(all(v)),v.end());}
template<class S,class T> ostream& operator<<(ostream& o,const pair<S,T> &p){
	return o<<"("<<p.fs<<","<<p.sc<<")";
}
template<class T> ostream& operator<<(ostream& o,const vector<T> &vc){
	o<<"{";
	for(const T& v:vc) o<<v<<",";
	o<<"}";
	return o;
}
constexpr ll TEN(int n) { return (n == 0) ? 1 : 10 * TEN(n-1); }

#ifdef LOCAL
#define show(x) cerr << "LINE" << __LINE__ << " : " << #x << " = " << (x) << endl
void dmpr(ostream& os){os<<endl;}
template<class T,class... Args>
void dmpr(ostream&os,const T&t,const Args&... args){
	os<<t<<" ~ ";
	dmpr(os,args...);
}
#define shows(...) cerr << "LINE" << __LINE__ << " : ";dmpr(cerr,##__VA_ARGS__)
#define dump(x) cerr << "LINE" << __LINE__ << " : " << #x << " = {";  \
	for(auto v: x) cerr << v << ","; cerr << "}" << endl;
#else
#define show(x) void(0)
#define dump(x) void(0)
#define shows(...) void(0)
#endif

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);		//DON'T USE scanf/printf/puts !!
	cout << fixed << setprecision(20);

	string s; cin >> s;
	ll N = s.size();
	ll O = 0, Z = 0;
	rep(i,N){
		if(s[i] == '0') Z++;
		else O++;
	}
	ll ans = 0;
	rep(i,O) ans += (i+1+1)/2;

	ll o = 0;
	ll z = 0;
	int p0 = -1;
	rep(i,N+1){
		if(i == N || s[i] == '0'){
			int n = i-p0-1;
			ans += n/2 * Z;
			if(n%2==1){
				if(p0%2 == 1){	//happy
					ans += (z+1)/2 + Z-z;
				}else{
					ans += z/2 + Z-z;
				}
			}
			p0 = i;
			z++;
		}else{
			o++;
		}
	}
	cout << ans << endl;
}
