#include <bits/stdc++.h>
using namespace std;
struct Trie {
  struct Node {
    int child[2];
    int sum;
    Node() {
      sum = 0;
      memset(child, -1, sizeof(child));
    }
  };
  vector<Node> nodes;
  Trie() { nodes.push_back(Node()); }
  void insert(string s) {
    int v = 0;
    for (char c : s) {
      nodes[v].sum++;
      int p = c - '0';
      if ((nodes[v].child[p]) == -1) {
        (nodes[v].child[p]) = nodes.size();
        nodes.push_back(Node());
      }
      v = (nodes[v].child[p]);
    }
    nodes[v].sum++;
  }
  void insertAll(string s) {
    for (int i = 0; i < s.size(); ++i) insert(s.substr(i));
  }
  void print(int u = 0, int lvl = 0, int lim = 2) {
    if (u == -1) return;
    print((nodes[u].child[0]), lvl + 1);
    cout << string(lvl * 2, ' ') << nodes[u].sum << endl;
    print((nodes[u].child[1]), lvl + 1, lim);
  }
  void printPath(const string &s, int doMore = 1, int u = 0, int lvl = 0) {
    if (u == -1) return;
    int p = lvl < s.size() ? s[lvl] - '0' : -1;
    cout << string(lvl * 2, ' ') << nodes[u].sum << endl;
    if (doMore) {
      printPath(s, p == 0, (nodes[u].child[0]), lvl + 1);
      printPath(s, p == 1, (nodes[u].child[1]), lvl + 1);
    }
  }
};
struct Fibo {
  string pre, suf;
  long long sz;
  Fibo(const string &s) {
    pre = s;
    suf = s;
    sz = s.size();
  }
  Fibo operator+(const Fibo &o) const {
    Fibo r("");
    r.sz = sz + o.sz;
    r.pre = (pre + o.pre);
    r.suf = (suf + o.suf);
    if (r.pre.size() > 220) r.pre = r.pre.substr(0, 220);
    if (r.suf.size() > 220) r.suf = r.suf.substr(r.suf.size() - 220);
    if (r.sz >= 4000000000000000000LL) r.sz = 4000000000000000000LL;
    return r;
  }
};
pair<long long, long long> countWith(string s, int N,
                                     const vector<Fibo> &fibos) {
  vector<vector<pair<long long, long long>>> P(
      max(N, 2), vector<pair<long long, long long>>(
                     s.size() + 1, pair<long long, long long>(0, 0)));
  P[0][0].first = 1;
  P[1][0].second = 1;
  for (int i = 2; i < N; ++i) {
    string suf = fibos[i - 2].suf;
    string pre = fibos[i - 1].pre;
    P[i] = P[i - 1];
    P[i][s.size()].first = min(4000000000000000000LL,
                               P[i][s.size()].first + P[i - 2][s.size()].first);
    P[i][s.size()].second =
        min(4000000000000000000LL,
            P[i][s.size()].second + P[i - 2][s.size()].second);
    for (int sz = 0; sz < suf.size(); ++sz) {
      string rsuf = suf.substr(sz);
      int p = 0;
      while (p < s.size() && p < rsuf.size() && s[p] == rsuf[p]) p++;
      if (p < suf.size() - sz) continue;
      rsuf += pre;
      while (p < s.size() && p < rsuf.size() && s[p] == rsuf[p]) p++;
      if (p != rsuf.size()) {
        pair<long long, long long> &t = P[i][p];
        long long &d = (rsuf[p] == '0' ? t.first : t.second);
        d = min(d + 1, 4000000000000000000LL);
      }
    }
  }
  return P[N - 1][s.size()];
}
string findKth(long long K, int N, const vector<Fibo> &fibos, int ML = 200) {
  string s;
  string me = fibos[N - 1].suf;
  while (K > 0 && s.size() < ML) {
    pair<long long, long long> cn = countWith(s, N, fibos);
    if (s.size() && me.substr(me.size() - s.size()) == s) {
      K--;
      if (K == 0) break;
    }
    if (K <= cn.first) {
      s += "0";
    } else {
      K -= cn.first;
      s += "1";
    }
  }
  return s;
}
int main() {
  vector<Fibo> fibos;
  Fibo a("0");
  Fibo b("1");
  fibos.push_back(a);
  fibos.push_back(b);
  int N = 210;
  for (int i = 3; i <= N; ++i) {
    Fibo c = a + b;
    a = b;
    b = c;
    fibos.push_back(c);
  }
  long long K, M;
  cin >> N >> K >> M;
  cout << findKth(K, N + 1, fibos, M) << endl;
}

### Prompt

In cpp, your task is to solve the following problem:

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;
struct Trie {
  struct Node {
    int child[2];
    int sum;
    Node() {
      sum = 0;
      memset(child, -1, sizeof(child));
    }
  };
  vector<Node> nodes;
  Trie() { nodes.push_back(Node()); }
  void insert(string s) {
    int v = 0;
    for (char c : s) {
      nodes[v].sum++;
      int p = c - '0';
      if ((nodes[v].child[p]) == -1) {
        (nodes[v].child[p]) = nodes.size();
        nodes.push_back(Node());
      }
      v = (nodes[v].child[p]);
    }
    nodes[v].sum++;
  }
  void insertAll(string s) {
    for (int i = 0; i < s.size(); ++i) insert(s.substr(i));
  }
  void print(int u = 0, int lvl = 0, int lim = 2) {
    if (u == -1) return;
    print((nodes[u].child[0]), lvl + 1);
    cout << string(lvl * 2, ' ') << nodes[u].sum << endl;
    print((nodes[u].child[1]), lvl + 1, lim);
  }
  void printPath(const string &s, int doMore = 1, int u = 0, int lvl = 0) {
    if (u == -1) return;
    int p = lvl < s.size() ? s[lvl] - '0' : -1;
    cout << string(lvl * 2, ' ') << nodes[u].sum << endl;
    if (doMore) {
      printPath(s, p == 0, (nodes[u].child[0]), lvl + 1);
      printPath(s, p == 1, (nodes[u].child[1]), lvl + 1);
    }
  }
};
struct Fibo {
  string pre, suf;
  long long sz;
  Fibo(const string &s) {
    pre = s;
    suf = s;
    sz = s.size();
  }
  Fibo operator+(const Fibo &o) const {
    Fibo r("");
    r.sz = sz + o.sz;
    r.pre = (pre + o.pre);
    r.suf = (suf + o.suf);
    if (r.pre.size() > 220) r.pre = r.pre.substr(0, 220);
    if (r.suf.size() > 220) r.suf = r.suf.substr(r.suf.size() - 220);
    if (r.sz >= 4000000000000000000LL) r.sz = 4000000000000000000LL;
    return r;
  }
};
pair<long long, long long> countWith(string s, int N,
                                     const vector<Fibo> &fibos) {
  vector<vector<pair<long long, long long>>> P(
      max(N, 2), vector<pair<long long, long long>>(
                     s.size() + 1, pair<long long, long long>(0, 0)));
  P[0][0].first = 1;
  P[1][0].second = 1;
  for (int i = 2; i < N; ++i) {
    string suf = fibos[i - 2].suf;
    string pre = fibos[i - 1].pre;
    P[i] = P[i - 1];
    P[i][s.size()].first = min(4000000000000000000LL,
                               P[i][s.size()].first + P[i - 2][s.size()].first);
    P[i][s.size()].second =
        min(4000000000000000000LL,
            P[i][s.size()].second + P[i - 2][s.size()].second);
    for (int sz = 0; sz < suf.size(); ++sz) {
      string rsuf = suf.substr(sz);
      int p = 0;
      while (p < s.size() && p < rsuf.size() && s[p] == rsuf[p]) p++;
      if (p < suf.size() - sz) continue;
      rsuf += pre;
      while (p < s.size() && p < rsuf.size() && s[p] == rsuf[p]) p++;
      if (p != rsuf.size()) {
        pair<long long, long long> &t = P[i][p];
        long long &d = (rsuf[p] == '0' ? t.first : t.second);
        d = min(d + 1, 4000000000000000000LL);
      }
    }
  }
  return P[N - 1][s.size()];
}
string findKth(long long K, int N, const vector<Fibo> &fibos, int ML = 200) {
  string s;
  string me = fibos[N - 1].suf;
  while (K > 0 && s.size() < ML) {
    pair<long long, long long> cn = countWith(s, N, fibos);
    if (s.size() && me.substr(me.size() - s.size()) == s) {
      K--;
      if (K == 0) break;
    }
    if (K <= cn.first) {
      s += "0";
    } else {
      K -= cn.first;
      s += "1";
    }
  }
  return s;
}
int main() {
  vector<Fibo> fibos;
  Fibo a("0");
  Fibo b("1");
  fibos.push_back(a);
  fibos.push_back(b);
  int N = 210;
  for (int i = 3; i <= N; ++i) {
    Fibo c = a + b;
    a = b;
    b = c;
    fibos.push_back(c);
  }
  long long K, M;
  cin >> N >> K >> M;
  cout << findKth(K, N + 1, fibos, M) << endl;
}

```