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
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>  
#include <complex>
#include <queue>
#include <set>
#include <unordered_set>
#include <list>
#include <chrono>
#include <random>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
#include <map>
#include <unordered_map>
#include <stack>
#include <iomanip>
#include <fstream>
 
using namespace std;
 
typedef long long ll;
typedef long double ld;
typedef pair<int,int> p32;
typedef pair<ll,ll> p64;
typedef pair<double,double> pdd;
typedef vector<ll> v64;
typedef vector<int> v32;
typedef vector<vector<int> > vv32;
typedef vector<vector<ll> > vv64;
typedef vector<vector<p64> > vvp64;
typedef vector<p64> vp64;
typedef vector<p32> vp32;
ll MOD = 1e9+7;
double eps = 1e-12;
#define forn(i,e) for(ll i = 0; i < e; i++)
#define forsn(i,s,e) for(ll i = s; i < e; i++)
#define rforn(i,s) for(ll i = s; i >= 0; i--)
#define rforsn(i,s,e) for(ll i = s; i >= e; i--)
#define ln "\n"
#define dbg(x) cout<<#x<<" = "<<x<<ln
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define INF 2e18
#define fast_cin() ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
#define all(x) (x).begin(), (x).end()
#define sz(x) ((ll)(x).size())
 
ll mpow(ll a, ll b){
	if(a==0) return 0;
	if(b >= (MOD-1)) b %= (MOD-1);
	if(b==0) return 1;
	ll t1 = mpow(a,b/2);
	t1 *= t1;
	t1 %= MOD;
	if(b%2) t1 *= a;
	t1 %= MOD;
	return t1;
}
 
ll mpow(ll a, ll b, ll p){
	if(a==0) return 0;
	if(b==0) return 1;
	ll t1 = mpow(a,b/2,p);
	t1 *= t1;
	t1 %= p;
	if(b%2) t1 *= a;
	t1 %= p;
	return t1;
}
 
ll modinverse(ll a, ll m){
	ll m0 = m; 
    ll y = 0, x = 1; 
    if (m == 1) return 0; 
    while (a > 1){
        ll q = a / m; 
        ll t = m; 
        m = a % m, a = t; 
        t = y; 
        y = x - q * y; 
        x = t; 
    }
    if (x < 0) x += m0; 
    return x; 
}

mt19937_64 mt(chrono::steady_clock::now().time_since_epoch().count());
 
ll range(ll l, ll r){
	return l + mt()%(r-l+1);
}
 
ll rev(ll v){
	return mpow(v,MOD-2);
}

void solve(){
	string s;
	cin >> s;
	vp64 vec;
	ll ones = 0;
	forn(i,sz(s)){
		if(s[i]=='1') ones++;
		else vec.pb(mp(ones%2,i));
	}
	ll ans = 0;
	ll csz=0;
	ll cans = 0;
	forn(i,ones){
		csz++;
		if(csz%2) cans++;
		ans += cans;
	}
	ll cur = 1;
	ll rem = sz(vec);
	v64 odd;
	for(auto it : vec){
		if(it.fi==cur){
			cur^=1;
			csz++;
			if(csz%2) cans++;
			ans += cans;
			rem--;
		}
		else{
			odd.pb(it.se);
		}
	}
	ll in=0;
	ll parity=1;
	forn(i,sz(s)) {
		if(in<sz(odd) && odd[in]==i){
			in++;
		}
		else{
			if(s[i]=='1') {
				ans += parity*(rem-in);
				ans += (in+1-parity)/2;
			}
			parity ^= 1;
		}
	}
	cout << ans << ln;
}
 
int main()
{
	fast_cin();
    ll t=1;
    // cin >> t;
    forn(i,t) {
    	// cout << "Case #" << i+1 << ": ";
    	solve();
    }
    return 0;
}