#include <bits/stdc++.h>
using namespace std;
long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
const int maxn = 220;
int n, m;
long long k;
string s1[maxn], s2[maxn];
long long f[maxn];
bool pd(string a, string b) {
  int len = a.size();
  int l = b.size();
  return b.substr(l - len, len) == a;
}
int main() {
  n = read(), k = read(), m = read();
  s1[0] = s2[0] = "0";
  s1[1] = s2[1] = "1";
  for (register int i = (int)(2); i <= (int)(n); i++) {
    s1[i] = s1[i - 2];
    if (s1[i].size() < m) {
      s1[i] += s1[i - 1];
      if (s1[i].size() > m) s1[i] = s1[i].substr(0, m);
    }
    s2[i] = s2[i - 1];
    if (s2[i].size() < m) {
      s2[i] = s2[i - 2] + s2[i - 1];
      if (s2[i].size() > m) s2[i] = s2[i].substr(s2[i].size() - m, m);
    }
  }
  string ans;
  ans.clear();
  for (register int i = (int)(1); i <= (int)(m); i++) {
    memset(f, 0, sizeof(f));
    f[0] = 1;
    if (i != 1 && ans[0] == '1') f[1] = 1;
    for (register int j = (int)(2); j <= (int)(n); j++) {
      f[j] = f[j - 1] + f[j - 2];
      string nw = s2[j - 2] + s1[j - 1], t = ans + '1';
      for (int k = 0; k < s2[j - 2].size(); k++) {
        if (k + i <= s2[j - 2].size()) continue;
        string ss;
        if (k + i > nw.size())
          ss = nw.substr(k);
        else
          ss = nw.substr(k, i);
        int len = s2[j - 2].size() - k;
        if (ss >= t && s2[j - 2].substr(k, len) == t.substr(0, len)) f[j]--;
      }
      if (f[j] >= k) {
        f[n] = k;
        break;
      }
    }
    if (f[n] >= k)
      ans += '0';
    else
      ans += '1';
    if (f[n] + 1 == k && pd(ans, s2[n])) break;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
const int maxn = 220;
int n, m;
long long k;
string s1[maxn], s2[maxn];
long long f[maxn];
bool pd(string a, string b) {
  int len = a.size();
  int l = b.size();
  return b.substr(l - len, len) == a;
}
int main() {
  n = read(), k = read(), m = read();
  s1[0] = s2[0] = "0";
  s1[1] = s2[1] = "1";
  for (register int i = (int)(2); i <= (int)(n); i++) {
    s1[i] = s1[i - 2];
    if (s1[i].size() < m) {
      s1[i] += s1[i - 1];
      if (s1[i].size() > m) s1[i] = s1[i].substr(0, m);
    }
    s2[i] = s2[i - 1];
    if (s2[i].size() < m) {
      s2[i] = s2[i - 2] + s2[i - 1];
      if (s2[i].size() > m) s2[i] = s2[i].substr(s2[i].size() - m, m);
    }
  }
  string ans;
  ans.clear();
  for (register int i = (int)(1); i <= (int)(m); i++) {
    memset(f, 0, sizeof(f));
    f[0] = 1;
    if (i != 1 && ans[0] == '1') f[1] = 1;
    for (register int j = (int)(2); j <= (int)(n); j++) {
      f[j] = f[j - 1] + f[j - 2];
      string nw = s2[j - 2] + s1[j - 1], t = ans + '1';
      for (int k = 0; k < s2[j - 2].size(); k++) {
        if (k + i <= s2[j - 2].size()) continue;
        string ss;
        if (k + i > nw.size())
          ss = nw.substr(k);
        else
          ss = nw.substr(k, i);
        int len = s2[j - 2].size() - k;
        if (ss >= t && s2[j - 2].substr(k, len) == t.substr(0, len)) f[j]--;
      }
      if (f[j] >= k) {
        f[n] = k;
        break;
      }
    }
    if (f[n] >= k)
      ans += '0';
    else
      ans += '1';
    if (f[n] + 1 == k && pd(ans, s2[n])) break;
  }
  cout << ans << endl;
  return 0;
}

```