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>
#define int long long
#define ll long long
using ull = unsigned long long;
using namespace std;
#define dump(x)                             \
    if (dbg) {                              \
        cerr << #x << " = " << (x) << endl; \
    }
#define overload4(_1, _2, _3, _4, name, ...) name
#define FOR1(n) for (ll i = 0; i < (n); ++i)
#define FOR2(i, n) for (ll i = 0; i < (n); ++i)
#define FOR3(i, a, b) for (ll i = (a); i < (b); ++i)
#define FOR4(i, a, b, c) for (ll i = (a); i < (b); i += (c))
#define FOR(...) overload4(__VA_ARGS__, FOR4, FOR3, FOR2, FOR1)(__VA_ARGS__)
#define FORR(i, a, b) for (int i = (a); i <= (b); ++i)
#define bit(n, k) ((n >> k) & 1) /*nのk bit目*/
namespace mydef {
const int INF = 1ll << 60;
const int MOD = 1e9 + 7;
template <class T>
bool chmin(T& a, const T& b) {
    if (a > b) {
        a = b;
        return 1;
    } else
        return 0;
}
template <class T>
bool chmax(T& a, const T& b) {
    if (a < b) {
        a = b;
        return 1;
    } else
        return 0;
}
void Yes(bool flag = true) {
    if (flag)
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
}
void No(bool flag = true) {
    Yes(!flag);
}
void YES(bool flag = true) {
    if (flag)
        cout << "YES" << endl;
    else
        cout << "NO" << endl;
}
void NO(bool flag = true) {
    YES(!flag);
}
template <typename A, size_t N, typename T>
void Fill(A (&array)[N], const T& val) {
    std::fill((T*)array, (T*)(array + N), val);
}
bool dbg = false;
}  // namespace mydef
using namespace mydef;
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define lb lower_bound
#define ub upper_bound
#define all(v) (v).begin(), (v).end()
#define SZ(x) ((int)(x).size())
#define vi vector<int>
#define vvi vector<vector<int>>
#define vp vector<pair<int, int>>
#define vvp vector<vector<pair<int, int>>>
#define pi pair<int, int>
//#define P pair<int, int>
//#define V vector<int>
//#define S set<int>
#define asn ans

string T;

void solve() {
    int cnt1 = 0;
    for (auto& c : T)
        cnt1 += c;

    vp V;
    int tmp_o = 0, tmp_e = 0;
    int now = 1;


    for (int i = 0; i < T.size(); i++) {
        if (V.size()) {
            if (T[i]) {
                if (now % 2 == 1) {
                    V[V.size() - 1].first++;
                } else {
                    V[V.size() - 1].second++;
                }
                now = !now;
            } else {
                V.push_back({0, 0});
            }
        } else {
            if (T[i]) {
                if (now % 2 == 1) {
                    tmp_o++;
                } else {
                    tmp_e++;
                }
                now = !now;
            } else {
                V.push_back({0, 0});
            }
        }
    }

    /*
    for (int i = T.size() - 1; i >= 0; i--) {
        if (T[i]) {
            if (i % 2 == 0)
                tmp_o++;
            else
                tmp_e++;
        } else {
            V.push_back({tmp_o, tmp_e});
            tmp_e = 0;
            tmp_o = 0;
        }
    }
    */

    int ans = 0;
    for (int i = 1; i <= cnt1; i++) {
        ans += (i + 1) / 2;
    }
    cerr << ans << endl;
    ans += tmp_o * (int)V.size();
    cerr << ans << endl;

    int N = V.size();

    for (int i = 0; i < V.size(); i++) {
        int n = i + 1;
        //n回ひっくり返す

        int odd, even;  //何回それを取るか
        //cerr << i << " " << V[i].first << " " << V[i].second << endl;

        if (V[i].first > V[i].second) {
            if (n % 2 == 0) {
                even = n / 2;
                odd = N - even;
            } else {
                even = (n + 1) / 2;
                odd = N - even;
            }
        } else {
            if (n % 2 == 0) {
                odd = n / 2;
                even = N - odd;
            } else {
                odd = (n - 1) / 2;
                even = N - odd;
            }
        }

        ans += V[i].first * odd + V[i].second * even;
    }

    cout << ans << endl;
}

signed main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);

    cin >> T;
    for (auto& c : T)
        c -= '0';

    solve();
    return 0;
}
