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;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
mt19937 rnf(2106);
long long l, r, k;
void solv() {
  cin >> l >> r >> k;
  if (r - l + 1 <= 4) {
    long long ans = l;
    vector<long long> vans;
    vans.push_back(l);
    for (long long x = 0; x < (1 << (r - l + 1)); ++x) {
      long long u = 0;
      vector<long long> v;
      for (long long i = 0; i < 4; ++i) {
        if ((x & (1 << i))) {
          v.push_back(l + i);
          u ^= (l + i);
        }
      }
      if (1 <= v.size() && v.size() <= k) {
        if (u < ans) {
          ans = u;
          vans = v;
        }
      }
    }
    cout << ans << endl;
    cout << vans.size() << endl;
    for (int i = 0; i < vans.size(); ++i) cout << vans[i] << ' ';
    cout << endl;
    return;
  }
  if (k == 1) {
    cout << l << endl;
    cout << 1 << endl;
    cout << l << endl;
    return;
  }
  if (k == 2) {
    cout << 1 << endl;
    if (l % 2 == 0) {
      cout << 2 << endl;
      cout << l << ' ' << l + 1 << endl;
    } else {
      cout << 2 << endl;
      cout << l + 1 << ' ' << l + 2 << endl;
    }
    return;
  }
  if (k >= 4) {
    cout << 0 << endl;
    if (l % 2 == 0) {
      cout << 4 << endl;
      cout << l << ' ' << l + 1 << ' ' << l + 2 << ' ' << l + 3 << endl;
    } else {
      cout << 4 << endl;
      cout << l + 1 << ' ' << l + 2 << ' ' << l + 3 << ' ' << l + 4 << endl;
    }
    return;
  }
  for (long long i = 0; i < 58; ++i) {
    long long x = (1LL << i) | (1LL << (i + 1));
    long long y = (1LL << (i + 1)) | ((1LL << i) - 1);
    long long z = (1LL << i) | ((1LL << i) - 1);
    if (l <= z && x <= r) {
      cout << 0 << endl;
      cout << 3 << endl;
      cout << x << ' ' << y << ' ' << z << endl;
      return;
    }
  }
  cout << 1 << endl;
  if (l % 2 == 0) {
    cout << 2 << endl;
    cout << l << ' ' << l + 1 << endl;
  } else {
    cout << 2 << endl;
    cout << l + 1 << ' ' << l + 2 << endl;
  }
}
int main() {
  solv();
  return 0;
}