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 <iostream>
#include <string>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <map>
#include <set>
#include <algorithm>
#include <queue>
#include <stack>
#include <functional>
#include <bitset>
#include <assert.h>
using namespace std;
typedef long long ll;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<char> vc;
typedef vector<string> vs;
typedef vector<bool> vb;
typedef vector<double> vd;
typedef pair<ll,ll> P;
typedef vector<P> vpl;
typedef tuple<ll,ll,ll> tapu;
#define rep(i,n) for(ll i=0; i<(n); i++)
#define REP(i,a,b) for(int i=(a); i<(b); i++)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
const int inf = 1<<30;
const ll linf = 1LL<<62;
const int MAX = 510000;
ll dy[8] = {1,-1,0,0,1,-1,1,-1};
ll dx[8] = {0,0,1,-1,1,-1,-1,1};
const double pi = acos(-1);
const double eps = 1e-7;
template<typename T1,typename T2> inline bool chmin(T1 &a,T2 b){
	if(a>b){
		a = b; return true;
	}
	else return false;
}
template<typename T1,typename T2> inline bool chmax(T1 &a,T2 b){
	if(a<b){
		a = b; return true;
	}
	else return false;
}
template<typename T> inline void print(T &a){
    rep(i,a.size()) cout << a[i] << " ";
    cout << "\n";
}
template<typename T1,typename T2> inline void print2(T1 a, T2 b){cout << a << " " << b << "\n";}
template<typename T1,typename T2,typename T3> inline void print3(T1 a, T2 b, T3 c){
	cout << a << " " << b << " " << c << "\n";
}
const int mod = 1e9 + 7;
//const int mod = 998244353;

int main(){
	string s; cin >> s;
	ll n = s.size();
	ll zerosum = 0;
	ll ans = 0;
	rep(i,n){
		if(s[i] == '0') zerosum++;
	}
	ll zero = 0;
	rep(i,n){
		if(s[i] == '1'){
			ll j = i;
			while(j < n && s[j] == '1') j++;
			ll num = j - i;
			if(num % 2 == 0){
				ans += zerosum * (num / 2);
			}else{
				ans += (zerosum - zero) * ((num+1) / 2);
				ans += zero * (num / 2);
				if(i & 1) ans += zero / 2;
				else ans += (zero + 1) / 2; 
			}
			i = j-1;
		}else{
			zero++;
		}
	}
	ll rest = n - zerosum;
	if(rest & 1) ans += (rest+1) / 2;
	rest /= 2;
	ans += rest * (rest + 1);
	cout << ans << "\n";
}