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;
long long n, m;
long long k;
string str[17], suf[2010];
long long f[250], d[250], h[250], qy[2][2];
string p[2], s[2];
string ans;
long long cmp(string alpha, string beta) {
  for (long long i = 0; i < min(alpha.length(), beta.length()); i++) {
    if (alpha[i] == beta[i]) continue;
    if (alpha[i] < beta[i])
      return 1;
    else
      return 0;
  }
  return alpha.length() < beta.length();
}
long long Try(string a) {
  long long res = 0;
  for (long long i = 0; i < a.length(); i++) {
    if (i + ans.length() - 1 >= a.length()) break;
    long long flag = 1;
    for (long long j = 0; j < ans.length(); j++) {
      if (a[i + j] != ans[j]) flag = 0;
    }
    res += flag;
  }
  return res;
}
long long solve() {
  long long len = ans.length();
  memset(d, 0, sizeof(d));
  memset(h, 0, sizeof(h));
  memset(qy, 0, sizeof(qy));
  p[0].clear();
  p[1].clear();
  s[0].clear();
  s[1].clear();
  d[14] = h[14] = 0;
  d[15] = h[15] = 1;
  for (long long i = 0; i < len - 1; i++)
    p[0].push_back(str[14][i]), p[1].push_back(str[15][i]);
  for (long long i = 1; i <= len - 1; i++)
    s[0].push_back(str[14][str[14].length() - len + i]);
  for (long long i = 1; i <= len - 1; i++)
    s[1].push_back(str[15][str[15].length() - len + i]);
  for (long long i = 0; i < 2; i++)
    for (long long j = 0; j < 2; j++) qy[i][j] = Try(p[i] + s[j]);
  f[14] = Try(str[14]);
  f[15] = Try(str[15]);
  for (long long i = 16; i <= n; i++) {
    h[i] = h[i - 1];
    d[i] = d[i - 2];
    f[i] = f[i - 1] + f[i - 2];
    f[i] += qy[d[i]][h[i]];
    if (f[i] > 1e18) return f[i];
  }
  return f[n];
}
int main() {
  scanf("%d%lld%d", &n, &k, &m);
  str[0] = "0";
  str[1] = "1";
  for (long long i = 2; i <= 15; i++) str[i] = str[i - 2] + str[i - 1];
  if (n <= 15) {
    for (long long i = 0; i < str[n].length(); i++) {
      for (long long j = i; j < str[n].length(); j++)
        suf[i + 1].push_back(str[n][j]);
    }
    sort(suf + 1, suf + str[n].length() + 1, cmp);
    for (long long i = 0; i < min(m, (long long)suf[k].length()); i++)
      printf("%c", suf[k][i]);
    return 0;
  } else {
    for (long long i = 1; i <= m; i++) {
      ans.push_back('0');
      long long z = solve();
      if (z < k) {
        k -= z;
        ans.pop_back();
        ans.push_back('1');
      }
      int flag = 1;
      for (int i = suf[n].length() - ans.length(), j = 0; i < suf[n].length();
           i++) {
        if (ans[j] != suf[n][i]) flag = false;
        j++;
      }
      k -= flag;
      if (k == 0) {
        for (int i = 0; i < ans.length(); i++) printf("%c", ans[i]);
        return 0;
      }
    }
    for (long long i = 0; i < ans.length(); i++) printf("%c", ans[i]);
  }
  return 0;
}
