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;
#define ll long long
#define ull unsigned long long
#define i128 __int128
#define mp make_pair
#define ld long double
typedef pair<int, int> pii;
typedef priority_queue<int, vector<int>, greater<int> > small_heap;
typedef priority_queue<int> big_heap;
const int N = 1e6 + 100;
char s[N];
int t[N], f[N];

int main() {
    scanf("%s", s + 1);
    int n = strlen(s + 1);
    int tot = 0, cnt[2] = {0, 0};
    ll ans = 0;
    for (int i = 1; i <= n; i++)
        tot += s[i] == '0';
    for (int i = 1; i <= n; i++)
        if (s[i] == '0') cnt[0]++;
        else {
            if (s[i + 1] == '1') {
                ans += tot + 1, i++;
                cnt[1] += 2;
            } else {
                cnt[1] += 1;
                ans += i % 2;
                ans += tot - cnt[0];
                if (i % 2) ans += cnt[0] / 2;
                else ans += (cnt[0] + 1) / 2;
            }
        }
    for (int i = 1; i <= cnt[1] - 1; i++)
        ans += (i + 1) / 2;
    cout << ans << endl;
    return 0;
}