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 optimize ("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

#pragma warning(disable:4786)
#pragma warning(disable:4996)
#include <ctime>
#include<list>
#include <numeric>
#include<bitset>
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<set>
#include<map>
#include<functional>
#include<string>
#include<cstring>
#include<cstdlib>
#include<queue>
#include<utility>
#include<fstream>
#include<sstream>
#include<cmath>
#include<stack>
#include<assert.h>
#include<unordered_map>
#include<unordered_set>
#include <array>
using namespace std;

#define MEM(a, b) memset(a, (b), sizeof(a))
#define CLR(a) memset(a, 0, sizeof(a))
#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define MIN(a, b) ((a) < (b) ? (a) : (b))
#define ABS(X) ( (X) > 0 ? (X) : ( -(X) ) )
#define S(X) ( (X) * (X) )
#define SZ(V) (int )V.size()
#define FORN(i, n) for(int i = 0; i < n; i++)
#define FORAB(i, a, b) for(int i = a; i <= b; i++)
#define ALL(V) V.begin(), V.end()
#define IN(A, B, C)  ((B) <= (A) && (A) <= (C))
#define AIN(A, B, C) assert(IN(A, B, C))

typedef long long int LL;
//typedef __int128 LLL;
typedef long long LLL;

typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
typedef pair<double, double> PDD;
typedef vector<int> VI;
typedef vector<LL> VL;
typedef vector<PLL> VPL;
typedef vector<PII> VP;
typedef vector<double> VD;
typedef vector<vector<int>> VVI;
typedef vector<string> VS;
typedef long double ld;

#define MAXN 1000006
//#define MAXN 1006

const LL MOD = 1000000007;
//const LL MOD = 998244353;
const LL INF = 2000000000000000001LL; //2e18 + 1

void solve(int ks) {
	char S[200005];
	scanf("%s", S);
	int cnt0 = 0, cnt1 = 0;
	for (int i = 0; S[i]; i++) {
		cnt0 += (S[i] == '0');
		cnt1 += (S[i] == '1');
	}

	LL ans = 0;
	// 0-0, 1-1, 2-2, 3-4, 4-6, 5-9, 6-12
	LL cur = 0, del = 1;
	for (int i = 1; i <= cnt1; i++) {
		cur += del;
		ans = cur;
		if (i % 2 == 0) {
			del++;
		}
	}

	VP V;
	for (int i = 0; S[i]; i++) {
		if (V.empty()) {
			V.push_back({ 1, S[i] - '0' });
		}
		else if (V.back().second == S[i] - '0') {
			V.back().first++;
			if (V.back().second) {
				V.pop_back();
				ans += cnt0;
			}
		}
		else V.push_back({ 1, S[i] - '0' });
	}

	int cur_cnt0 = 0;
	int pos = -1;
	for (int i = 0; i < SZ(V); i++) {
		if (V[i].second == 0) {
			cur_cnt0 += V[i].first;
			pos += V[i].first;
			continue;
		}
		pos++;
		int rem_cnt0 = cnt0 - cur_cnt0;
		if (pos % 2) {
			// even
			ans += rem_cnt0 + cur_cnt0 / 2;
		}
		else {
			// odd
			ans += rem_cnt0 + (cur_cnt0 + 1) / 2;
		}

	}
	printf("%lld\n", ans);
}

void gen() {
}

int main()
{
	double start_time = clock();
#ifdef LOCAL
	freopen("C:\\Home\\ContestCodes\\sample.in", "r", stdin);
	// freopen("C:\\Home\\ContestCodes\\sample.in", "r", stdin);
	//	freopen("out.out", "w", stdout);
#endif

	gen();

	if (0) {
		int T;
		scanf("%d", &T);
		//AIN(T, 1, 5000);
		for (int ks = 1; ks <= T; ks++) {
			solve(ks);
			if (ks % 1 == 0) fprintf(stderr, "%d done\n", ks);
		}
	}
	else {
		solve(0);
	}

	double end_time = clock();
	fprintf(stderr, "Time = %lf\n", (end_time - start_time) / CLOCKS_PER_SEC);

	return 0;
}
