#include <bits/stdc++.h>
using namespace std;
namespace Base {
const int inf = 0x3f3f3f3f, INF = 0x7fffffff;
const long long infll = 0x3f3f3f3f3f3f3f3fll, INFll = 0x7fffffffffffffffll;
template <typename T>
void read(T &x) {
  x = 0;
  int fh = 1;
  double num = 1.0;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') fh = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  if (ch == '.') {
    ch = getchar();
    while (isdigit(ch)) {
      num /= 10;
      x = x + num * (ch - '0');
      ch = getchar();
    }
  }
  x = x * fh;
}
template <typename T>
void chmax(T &x, T y) {
  x = x < y ? y : x;
}
template <typename T>
void chmin(T &x, T y) {
  x = x > y ? y : x;
}
}  // namespace Base
using namespace Base;
const int N = 510;
bool tag[N];
int nxt[N], n, m, len[N], pre[N][N], bak[N][N], now[N], ch[N];
long long k, cnt[N];
void kmp(int *a, int lena, int *b, int lenb) {
  memset(tag, 0, sizeof(tag));
  for (int i = 1; i <= lenb; i++) {
    int p = i - 1;
    while (p != 0 && b[nxt[p] + 1] != b[i]) p = nxt[p];
    if (b[nxt[p] + 1] != b[i] || i == 1)
      nxt[i] = 0;
    else
      nxt[i] = nxt[p] + 1;
  }
  for (int i = 1, j = 1; i <= lena; i++) {
    while (j != 1 && a[i] != b[j]) j = nxt[j - 1] + 1;
    if (a[i] == b[j]) j++;
    if (j > lenb) {
      tag[i] = true, j = nxt[j - 1] + 1;
    }
  }
}
int main() {
  read(n);
  read(k);
  read(m);
  k = k + 1;
  len[0] = len[1] = 1;
  pre[0][1] = 0;
  pre[1][1] = 1;
  bak[0][1] = 0;
  bak[1][1] = 1;
  for (int j = 2; j <= n; j++) {
    for (int k = 1; k <= len[j - 2]; k++) pre[j][k] = pre[j - 2][k];
    if (len[j - 2] < m) {
      for (int k = len[j - 2] + 1; k <= min(len[j - 2] + len[j - 1], m); k++)
        pre[j][k] = pre[j - 1][k - len[j - 2]];
    }
    int tmp = 0;
    if (len[j - 1] < m) {
      tmp = min(len[j - 2], m - len[j - 1]);
      for (int k = 1; k <= tmp; k++)
        bak[j][k] = bak[j - 2][len[j - 2] - tmp + k];
    }
    for (int k = 1; k <= len[j - 1]; k++) bak[j][k + tmp] = bak[j - 1][k];
    len[j] = min(m, len[j - 1] + len[j - 2]);
  }
  for (int i = 1; i <= m; i++) {
    bool flag = false;
    for (int j = 1; j <= i - 1; j++)
      if (bak[n][len[n] - i + j + 1] != ch[j]) {
        flag = true;
        break;
      }
    if (flag == false) {
      k--;
      if (k == 0) {
        for (int j = 1; j <= i - 1; j++) printf("%d", ch[j]);
        printf("\n");
        exit(0);
      }
    }
    flag = false;
    ch[i] = 0;
    memset(cnt, 0, sizeof(cnt));
    if (i == 1) cnt[0] = 1;
    for (int j = 2; j <= n; j++) {
      cnt[j] = cnt[j - 1] + cnt[j - 2];
      if (cnt[j] >= k) {
        flag = true;
        break;
      }
      int id = 0, cut = 0;
      for (int k = 1; k <= len[j - 2]; k++) now[++id] = bak[j - 2][k];
      cut = id;
      for (int k = 1; k <= len[j - 1]; k++) now[++id] = pre[j - 1][k];
      kmp(now, id, ch, i);
      for (int k = cut + 1; k <= min(cut + i - 1, id); k++)
        if (tag[k] == true) cnt[j]++;
    }
    if (cnt[n] >= k || flag == true)
      ch[i] = 0;
    else
      ch[i] = 1, k -= cnt[n];
  }
  for (int i = 1; i <= m; i++) printf("%d", ch[i]);
  printf("\n");
  return 0;
}

### 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;
namespace Base {
const int inf = 0x3f3f3f3f, INF = 0x7fffffff;
const long long infll = 0x3f3f3f3f3f3f3f3fll, INFll = 0x7fffffffffffffffll;
template <typename T>
void read(T &x) {
  x = 0;
  int fh = 1;
  double num = 1.0;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') fh = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  if (ch == '.') {
    ch = getchar();
    while (isdigit(ch)) {
      num /= 10;
      x = x + num * (ch - '0');
      ch = getchar();
    }
  }
  x = x * fh;
}
template <typename T>
void chmax(T &x, T y) {
  x = x < y ? y : x;
}
template <typename T>
void chmin(T &x, T y) {
  x = x > y ? y : x;
}
}  // namespace Base
using namespace Base;
const int N = 510;
bool tag[N];
int nxt[N], n, m, len[N], pre[N][N], bak[N][N], now[N], ch[N];
long long k, cnt[N];
void kmp(int *a, int lena, int *b, int lenb) {
  memset(tag, 0, sizeof(tag));
  for (int i = 1; i <= lenb; i++) {
    int p = i - 1;
    while (p != 0 && b[nxt[p] + 1] != b[i]) p = nxt[p];
    if (b[nxt[p] + 1] != b[i] || i == 1)
      nxt[i] = 0;
    else
      nxt[i] = nxt[p] + 1;
  }
  for (int i = 1, j = 1; i <= lena; i++) {
    while (j != 1 && a[i] != b[j]) j = nxt[j - 1] + 1;
    if (a[i] == b[j]) j++;
    if (j > lenb) {
      tag[i] = true, j = nxt[j - 1] + 1;
    }
  }
}
int main() {
  read(n);
  read(k);
  read(m);
  k = k + 1;
  len[0] = len[1] = 1;
  pre[0][1] = 0;
  pre[1][1] = 1;
  bak[0][1] = 0;
  bak[1][1] = 1;
  for (int j = 2; j <= n; j++) {
    for (int k = 1; k <= len[j - 2]; k++) pre[j][k] = pre[j - 2][k];
    if (len[j - 2] < m) {
      for (int k = len[j - 2] + 1; k <= min(len[j - 2] + len[j - 1], m); k++)
        pre[j][k] = pre[j - 1][k - len[j - 2]];
    }
    int tmp = 0;
    if (len[j - 1] < m) {
      tmp = min(len[j - 2], m - len[j - 1]);
      for (int k = 1; k <= tmp; k++)
        bak[j][k] = bak[j - 2][len[j - 2] - tmp + k];
    }
    for (int k = 1; k <= len[j - 1]; k++) bak[j][k + tmp] = bak[j - 1][k];
    len[j] = min(m, len[j - 1] + len[j - 2]);
  }
  for (int i = 1; i <= m; i++) {
    bool flag = false;
    for (int j = 1; j <= i - 1; j++)
      if (bak[n][len[n] - i + j + 1] != ch[j]) {
        flag = true;
        break;
      }
    if (flag == false) {
      k--;
      if (k == 0) {
        for (int j = 1; j <= i - 1; j++) printf("%d", ch[j]);
        printf("\n");
        exit(0);
      }
    }
    flag = false;
    ch[i] = 0;
    memset(cnt, 0, sizeof(cnt));
    if (i == 1) cnt[0] = 1;
    for (int j = 2; j <= n; j++) {
      cnt[j] = cnt[j - 1] + cnt[j - 2];
      if (cnt[j] >= k) {
        flag = true;
        break;
      }
      int id = 0, cut = 0;
      for (int k = 1; k <= len[j - 2]; k++) now[++id] = bak[j - 2][k];
      cut = id;
      for (int k = 1; k <= len[j - 1]; k++) now[++id] = pre[j - 1][k];
      kmp(now, id, ch, i);
      for (int k = cut + 1; k <= min(cut + i - 1, id); k++)
        if (tag[k] == true) cnt[j]++;
    }
    if (cnt[n] >= k || flag == true)
      ch[i] = 0;
    else
      ch[i] = 1, k -= cnt[n];
  }
  for (int i = 1; i <= m; i++) printf("%d", ch[i]);
  printf("\n");
  return 0;
}

```