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
/* Generated by powerful Codeforces Tool
 * You can download the binary file in here https://github.com/xalanq/cf-tool (win, osx, linux)
 * Author: step_by_step
 * Time: 2020-03-19 17:35:03
**/

//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#pragma GCC optimize("unroll-loops")
  
#include <iostream>
#include <stdlib.h>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <deque>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <random>
#include <assert.h>
#include <memory.h>
#include <time.h>
#include <bitset>

#define uint unsigned int
#define ll long long
#define ull unsigned long long
#define ld long double
#define rep(i, l, r) for (int i = (l); i < (r); i++)
#define repb(i, r, l) for (int i = (r); i > (l); i--)
#define sz(a) (int)a.size()
#define fi first
#define se second
#define mp(a, b) make_pair(a, b)
#define rank qwertyuio
#define next dfghjk
#define prev fhsgfhjf
#define plus fsghsf
#define minus ytryr
  
using namespace std;
  
inline bool setmin(int &x, int y) { return (y < x) ? x = y, 1 : 0; }
inline bool setmax(int &x, int y) { return (y > x) ? x = y, 1 : 0; }
inline bool setmin(ll &x, ll y) { return (y < x) ? x = y, 1 : 0; }
inline bool setmax(ll &x, ll y) { return (y > x) ? x = y, 1 : 0; }
  
const int N = 100000;
const int inf = (int)1e9 + 1;
const ll big = (ll)1e18 + 1;
const int P = 239;
const int P1 = 31;
const int P2 = 57;
const int MOD = (int)1e9 + 7;
const int MOD1 = (int)1e9 + 9;
const int MOD2 = 998244353;
const ld eps = 1e-9;
const double pi = atan2(0, -1);
const int ABC = 26;

int main()
{
    //freopen("circus.in", "r", stdin);
    //freopen("circus.out", "w", stdout);
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.precision(20);
    cout << fixed;
    //ll TL = 10.95 * CLOCKS_PER_SEC;
    //clock_t time = clock();
    string s;
    cin >> s;
    ll ans = 0;
    int cnt0 = 0, cnt1 = 0;
    rep(i, 0, sz(s)) {
        if (s[i] == '0') {
            cnt0++;
        } else {
            cnt1++;
        }
    }
    repb(i, cnt1, 0) {
        ans += (i + 1) / 2;
    }
    string t;
    int cur = 0;
    rep(i, 0, sz(s)) {
        if (s[i] == '0') {
            t += '0';
        } else {
            cur++;
            if (i == sz(s) - 1 || s[i] != s[i + 1]) {
                ans += 1LL * cnt0 * (cur / 2);
                if (cur % 2 == 1) {
                    t += '1';
                }
                cur = 0;
            }
        }
    }
    int rem0 = 0;
    rep(i, 0, sz(t)) {
        if (t[i] == '0') {
            rem0++;
        }
    }
    int cur_1_pos = 0;
    rep(i, 0, sz(t)) {
        if (t[i] == '1') {
            int bad = (i + 1) / 2 - (cur_1_pos + 1) / 2;
            ans += rem0 - bad;
            cur_1_pos++;
        }
    }
    cout << ans << "\n";
    return 0;
}

