Problem: Little Victor adores the sets theory. Let us remind you that a set is a group of numbers where all numbers are pairwise distinct. Today Victor wants to find a set of integers S that has the following properties:

  * for all x <image> the following inequality holds l ≤ x ≤ r; 
  * 1 ≤ |S| ≤ k; 
  * lets denote the i-th element of the set S as si; value <image> must be as small as possible. 



Help Victor find the described set.

Input

The first line contains three space-separated integers l, r, k (1 ≤ l ≤ r ≤ 1012; 1 ≤ k ≤ min(106, r - l + 1)).

Output

Print the minimum possible value of f(S). Then print the cardinality of set |S|. Then print the elements of the set in any order.

If there are multiple optimal sets, you can print any of them.

Examples

Input

8 15 3


Output

1
2
10 11


Input

8 30 7


Output

0
5
14 9 28 11 16

Note

Operation <image> represents the operation of bitwise exclusive OR. In other words, it is the XOR operation.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
bool ara[50];
long long l, r, k;
int lefty;
bool two_power() {
  vector<long long> vv;
  for (int i = 0; i <= 45; i++) {
    if ((1LL << i) > r) {
      break;
    }
    if ((1LL << i) >= l) {
      vv.push_back((1LL << i));
    }
  }
  if (vv.size() < 2) {
    return 0;
  }
  if (vv[0] + vv[1] > r) {
    return 0;
  }
  cout << 0 << "\n";
  cout << 3 << "\n";
  cout << vv[0] << " " << vv[1] << " " << vv[0] + vv[1];
  return 1;
}
bool one_power() {
  if (ara[lefty]) {
    return 0;
  }
  if ((1LL << lefty) + (1LL << (lefty - 1)) > r) {
    return 0;
  }
  cout << 0 << "\n";
  cout << 3 << "\n";
  cout << l << " " << (1LL << lefty) + (1LL << (lefty - 1)) << " "
       << l + (1LL << (lefty)) - (1LL << (lefty - 1));
  return 1;
}
bool two_gap() {
  vector<long long> vv;
  for (int i = 0; i <= lefty; i++) {
    if (ara[i] == 0) {
      vv.push_back(i);
    }
  }
  if (vv[0] == 0) {
    if (vv.size() < 2) {
      return 0;
    }
    long long s1 = 0;
    for (int i = 45; i >= vv[1] + 1; i--) {
      if (ara[i]) {
        s1 += (1LL << i);
      }
    }
    s1 += (1LL << vv[1]);
    if (s1 + (1LL << vv[0]) > r) {
      return 0;
    }
    cout << 0 << "\n";
    cout << 4 << "\n";
    cout << l << " " << s1 + (1LL << vv[0]) << " " << s1 << " "
         << l + (1LL << vv[0]);
  } else if (vv[0] == 1) {
    if (vv.size() < 2) {
      return 0;
    }
    long long s1 = 0;
    for (int i = 45; i >= vv[1] + 1; i--) {
      if (ara[i]) {
        s1 += (1LL << i);
      }
    }
    long long s2 = s1;
    for (int i = vv[1] - 1; i >= vv[2] + 1; i--) {
      if (ara[i]) {
        s2 += (1LL << i);
      }
    }
    if ((s1 + (1LL << vv[1]) + 1LL) > r) {
      return 0;
    }
    cout << 0 << "\n";
    cout << 4 << "\n";
    cout << s1 + (1LL << vv[1]) << " " << s1 + (1LL << vv[1]) + 1LL << " "
         << s2 + (1LL << vv[0]) << " " << s2 + (1LL << vv[0]) + 1LL;
  } else {
    long long s1 = 0;
    for (int i = 45; i >= vv[0] + 1; i--) {
      if (ara[i]) {
        s1 += (1LL << i);
      }
    }
    if (s1 + (1LL << vv[0]) + 3LL > r) {
      return 0;
    }
    cout << 0 << "\n";
    cout << 4 << "\n";
    cout << s1 + (1LL << vv[0]) << " " << s1 + (1LL << vv[0]) + 1LL << " "
         << s1 + (1LL << vv[0]) + 2LL << " " << s1 + (1LL << vv[0]) + 3LL;
  }
  return 1;
}
bool one_gap() {
  if (l % 2 == 0) {
    cout << 1 << "\n";
    cout << 2 << "\n";
    cout << l << " " << l + 1;
    return 1;
  }
  if (r < (l + 2)) {
    return 0;
  }
  int kh = 0;
  for (int i = 0; i <= lefty; i++) {
    if (ara[i] == 0) {
      kh = i;
      break;
    }
  }
  long long yo = 0;
  for (int i = 45; i >= kh + 1; i--) {
    if (ara[i]) {
      yo += (1LL << i);
    }
  }
  yo += (1LL << kh);
  cout << 1 << "\n";
  cout << 2 << "\n";
  cout << yo << " " << yo + 1;
  return 1;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  ;
  cin >> l >> r >> k;
  if (k == 1) {
    cout << l << "\n";
    cout << 1 << "\n";
    cout << l << "\n";
    return 0;
  }
  long long yo = l;
  for (int i = 45; i >= 0; i--) {
    if ((1LL << i) <= yo) {
      ara[i] = 1;
      yo -= (1LL << i);
    }
  }
  for (int i = 45; i >= 0; i--) {
    if ((1LL << i) <= r) {
      lefty = i;
      break;
    }
  }
  if ((k >= 3) && two_power()) {
    return 0;
  }
  if ((k >= 3) && one_power()) {
    return 0;
  }
  if ((k >= 4) && (two_gap())) {
    return 0;
  }
  if ((k >= 2) && (one_gap())) {
    return 0;
  }
  if ((l ^ (l + 1LL)) < l) {
    cout << (l ^ (l + 1LL)) << "\n";
    cout << 2 << "\n";
    cout << l << " " << l + 1LL;
    return 0;
  }
  cout << l << "\n";
  cout << 1 << "\n";
  cout << l << "\n";
  return 0;
}