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;
int GCD(int a, int b) {
  if (!a) return b;
  return GCD(b % a, a);
}
int main() {
  clock_t startTime = clock();
  ios_base::sync_with_stdio(false);
  long long l, r, k;
  cin >> l >> r >> k;
  if (r - l + 1 <= 16) {
    cerr << "case1\n";
    vector<long long> ar, best;
    long long fbest = r + 1;
    for (long long kk = l; kk <= r; kk++) ar.push_back(kk);
    for (int i = (1); i < (int)(1 << (r - l + 1)); i++) {
      long long f = 0;
      vector<long long> cur;
      for (int j = (0); j < (int)((r - l + 1)); j++) {
        if (i & (1 << j)) {
          f = f ^ ar[j];
          cur.push_back(ar[j]);
        }
      }
      if (f < fbest && cur.size() <= k) {
        fbest = f;
        best = cur;
      }
    }
    cout << fbest << "\n";
    cout << best.size() << "\n";
    for (int i = (0); i < (int)(best.size()); i++) cout << best[i] << " ";
  } else {
    cerr << "case2\n";
    if (k >= 4) {
      cout << 0 << "\n";
      cout << 4 << "\n";
      for (long long last = r; last - 3 >= l; last--) {
        if ((last ^ (last - 1) ^ (last - 2) ^ (last - 3)) == 0) {
          cout << last << " " << last - 1 << " " << last - 2 << " " << last - 3
               << "\n";
          break;
        }
      }
    } else if (k == 3) {
      bool pos = false;
      long long a, b, c;
      for (int i = (1); i < (int)(61); i++) {
        a = (1LL << i) - 1;
        b = (1LL << i) + (((i - 1) >= 0) ? ((1LL << (i - 1)) - 1) : (0));
        c = (1LL << i) + (1LL << (i - 1));
        if ((a >= l && a <= r) && (b >= l && b <= r) && (c >= l && c <= r)) {
          pos = true;
          break;
        }
      }
      if (pos) {
        cout << "0\n3\n";
        cout << a << " " << b << " " << c << "\n";
      } else {
        cout << "1\n2\n";
        for (long long last = r; last - 1 >= l; last--) {
          if ((last ^ (last - 1)) == 1) {
            cout << last << " " << last - 1 << "\n";
            break;
          }
        }
      }
    } else if (k == 2) {
      cout << "1\n2\n";
      for (long long last = r; last - 1 >= l; last--) {
        if ((last ^ (last - 1)) == 1) {
          cout << last << " " << last - 1 << "\n";
          break;
        }
      }
    } else {
      cout << l << "\n1\n" << l << "\n";
    }
  }
  clock_t endTime = clock();
  cerr << "\nTime:" << double(endTime - startTime) / CLOCKS_PER_SEC
       << " seconds\n";
  return 0;
}