Let's denote (yet again) the sequence of Fibonacci strings:

F(0) =  0, F(1) =  1, F(i) = F(i - 2) + F(i - 1), where the plus sign denotes the concatenation of two strings.

Let's denote the lexicographically sorted sequence of suffixes of string F(i) as A(F(i)). For example, F(4) is 01101, and A(F(4)) is the following sequence: 01, 01101, 1, 101, 1101. Elements in this sequence are numbered from 1.

Your task is to print m first characters of k-th element of A(F(n)). If there are less than m characters in this suffix, then output the whole suffix.

Input

The only line of the input contains three numbers n, k and m (1 ≤ n, m ≤ 200, 1 ≤ k ≤ 10^{18}) denoting the index of the Fibonacci string you have to consider, the index of the element of A(F(n)) and the number of characters you have to output, respectively.

It is guaranteed that k does not exceed the length of F(n).

Output

Output m first characters of k-th element of A(F(n)), or the whole element if its length is less than m.

Examples

Input

4 5 3


Output

110


Input

4 3 3


Output

1
#include <bits/stdc++.h>
using namespace std;
const int Maxn = 205;
string pre1, pre2, suf, now, str[Maxn];
int n, m;
long long k, cnt[Maxn];
namespace brute_forces {
int N, M, first[1000005], second[1000005], buc[1000005], sa[1000005];
void radix_sort(int first[], int second[]) {
  for (int i = 0; i <= M; i++) buc[i] = 0;
  for (int i = 0; i < N; i++) buc[first[second[i]]]++;
  for (int i = 1; i <= M; i++) buc[i] += buc[i - 1];
  for (int i = N - 1; i >= 0; i--) sa[--buc[first[second[i]]]] = second[i];
}
void work(void) {
  N = str[n].size();
  M = 2;
  for (int i = 0; i < N; i++)
    first[i] = str[n][i] - '0', second[i] = i, sa[i] = 0;
  radix_sort(first, second);
  for (int w = 1; w <= N; w <<= 1) {
    int cnt = 0;
    for (int i = N - w; i < N; i++) second[cnt++] = i;
    for (int i = 0; i < N; i++)
      if (sa[i] >= w) second[cnt++] = sa[i] - w;
    radix_sort(first, second);
    swap(first, second);
    cnt = 0;
    first[sa[0]] = cnt++;
    for (int i = 1; i < N; i++)
      first[sa[i]] = (sa[i] + w < N && sa[i - 1] + w < N &&
                      second[sa[i]] == second[sa[i - 1]] &&
                      second[sa[i] + w] == second[sa[i - 1] + w])
                         ? cnt - 1
                         : cnt++;
    if (cnt >= N) break;
    M = cnt;
  }
}
}  // namespace brute_forces
int path(string x, string y, string z) {
  int tmp_ans = 0, sizZ = z.size(), sizX = x.size(), sizY = y.size();
  for (int i = 0; i < sizX; i++)
    if (sizX - i < sizZ && sizX - i + sizY >= sizZ) {
      for (int j = i; j < sizX; j++)
        if (x[j] != z[j - i]) {
          tmp_ans += x[j] < z[j - i];
          goto END;
        }
      for (int j = 0; j < sizZ - sizX + i; j++)
        if (y[j] != z[j + sizX - i]) {
          tmp_ans += y[j] < z[j + sizX - i];
          goto END;
        }
    END:;
    }
  return tmp_ans;
}
long long cal(void) {
  pre1 = pre2 = suf = "";
  int s = now.size();
  if (now.size() == 1)
    cnt[0] = 1;
  else
    cnt[0] = 0;
  if (now.size() == 2) pre2 = "0", pre1 = suf = "1";
  if (now.size() == 3) pre1 = suf = "1";
  for (int i = 2; i <= n; i++) {
    cnt[i] = cnt[i - 1] + cnt[i - 2];
    if (cnt[i] >= 1e18) return 1e18;
    if (i >= 15 || str[i - 2].size() >= now.size()) {
      if (i & 1)
        cnt[i] += path(suf, pre2, now);
      else
        cnt[i] += path(suf, pre1, now);
    } else
      cnt[i] += path(str[i - 2], str[i - 1], now);
    if (i & 1) {
      if ((int)pre1.size() == s) continue;
      int siz = str[i].size();
      suf = pre1 = "";
      for (int j = 0; j < min(s, (int)str[i].size()); j++) pre1 += str[i][j];
      for (int j = max(0, siz - s); j < siz; j++) suf += str[i][j];
    } else {
      if ((int)pre2.size() == s) continue;
      pre2 = "";
      for (int j = 0; j < min(s, (int)str[i].size()); j++) pre2 += str[i][j];
    }
  }
  return cnt[n];
}
int main() {
  scanf("%d%lld%d", &n, &k, &m);
  str[0] = "0", str[1] = "1";
  for (int i = 2; i <= 27; i++) str[i] = str[i - 2] + str[i - 1];
  if (n <= 27) {
    brute_forces::work();
    int siz = min((int)str[n].size() - brute_forces::sa[k - 1], m);
    for (int i = 0; i < siz; i++) putchar(str[n][brute_forces::sa[k - 1] + i]);
    return 0;
  }
  for (int i = 0; i < m; i++) {
    string tmp = now;
    now += '1';
    long long ans_now = cal();
    for (int j = 1; j < suf.size(); j++) {
      string tmpstr = "";
      for (int k = j; k < suf.size(); k++) tmpstr += suf[k];
      if (tmpstr < now) ans_now++;
    }
    if (ans_now == k - 1 && now == suf) {
      cout << now;
      return 0;
    }
    if (ans_now > k - 1) now = tmp, now += '0';
  }
  cout << now;
  return 0;
}
