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;


void __print(int x) {cerr << x;}
void __print(long x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(unsigned x) {cerr << x;}
void __print(unsigned long x) {cerr << x;}
void __print(unsigned long long x) {cerr << x;}
void __print(float x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '\"' << x << '\"';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}

template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? "," : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}
#ifndef ONLINE_JUDGE
#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define debug(x...)
#endif

#define rep(i, n)    for(int i = 0; i < (n); ++i)
#define repA(i, a, n)  for(int i = a; i <= (n); ++i)
#define repD(i, a, n)  for(int i = a; i >= (n); --i)
#define trav(a, x) for(auto& a : x)
#define all(x) x.begin(), x.end()
#define sz(x) (int)(x).size()
#define fill(a)  memset(a, 0, sizeof (a))
#define fst first
#define snd second
#define mp make_pair
#define pb push_back
typedef long double ld;
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;

void pre(){


}

void solve(){


}

int main() {
	cin.sync_with_stdio(0); cin.tie(0);
	cin.exceptions(cin.failbit);
	pre();
	string s;cin>>s;
	int cnt = 0,f=0;
	trav(i,s){
		if(i=='1') cnt++;
	}
	string t;
	trav(i,s){
		if(sz(t)&&t.back()=='1'&&i=='1'){
			t.pop_back();
			f++;
		}
		else t+=i;
	}
	ll ans = 0;
	rep(i,f) ans+=sz(s)-2*i;
	int cur = 0;
	vi v;
	trav(i,t){
		if(i=='0') cur++;
		else v.pb(cur),cur=0;
	}
	v.pb(cur);
	int n = sz(t);
	int sum  =0;
	rep(i,sz(v)-1){
		sum+=v[i];
		ans+=n-2*i;
		ans-=(sum+1)/2;
		ans+=i;
		sum++;
	}
	cout<<ans;
	return 0;
}

