#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>

using namespace std;

const int N = 300010;

string ans;
int n, f[N];

int main() {
    // freopen("mess.in", "r", stdin);
    // freopen("mess.out", "w", stdout);

    scanf("%d", &n);
    memset(f, 0x3f, sizeof(f));
    f[0] = 0;
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= sqrt(i); ++j) {
            f[i] = min(f[i], j + f[i - j * j]);
        }
    }

    int len = f[n], s = 0; // 需要拼接的长度，新的串的首位的值
    bool odd_ok = true;
    while (true) {
        if (!len) break;
        if (odd_ok) {
            for (int i = 1; i <= len && i * i <= n; i += 2) {
                if (f[n - i * i] + i == len) {
                    len -= i;
                    n -= i * i;
                    for (int j = 1; j <= i; ++j) {
                        if (j % 2 == 1) ans += '0';
                        else ans += '1';
                    }
                    break;
                }
                if (i >= min(len, (int) sqrt(n)) - 2) odd_ok = false;
            }
        }
        if (!odd_ok) {
            if (s == 0) {
                for (int i = min(len, (int) sqrt(n)) / 2 * 2; i >= 2; i -= 2) {
                    if (f[n - i * i] + i == len) {
                        len -= i;
                        n -= i * i;
                        for (int j = 1; j <= i; ++j) {
                            if (j % 2 == 1) ans += '0';
                            else ans += '1';
                        }
                        s = 1;
                        break;
                    }
                }
            } else {
                for (int i = 2; i <= min(len, (int) sqrt(n)) / 2 * 2; i += 2) {
                    if (f[n - i * i] + i == len) {
                        len -= i;
                        n -= i * i;
                        for (int j = 1; j <= i; ++j) {
                            if (j % 2 == 1) ans += '1';
                            else ans += '0';
                        }
                        s = 0;
                        break;
                    }
                }
            }
        }
    }

    cout << ans << endl;

    return 0;
}