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;
const int inf = 1e9;
const double eps = 1e-9;
long long mn;
vector<bool> mnbm;
long long msb(long long x) {
  long long ans = 1LL << 60;
  while (!(ans & x)) ans >>= 1;
  return ans;
}
void brute(int n, int k, vector<bool>& bm, const vector<long long>& v) {
  if (n == int(v.size())) {
    long long ans = 0;
    int bits = 0;
    for (int i = 0; i < int(v.size()); ++i)
      if (bm[i]) {
        ans ^= v[i];
        ++bits;
      }
    if (bits and bits <= k and ans < mn) {
      mn = ans;
      mnbm = bm;
    }
  } else {
    bm[n] = true;
    brute(n + 1, k, bm, v);
    bm[n] = false;
    brute(n + 1, k, bm, v);
  }
}
int main() {
  mn = 1LL << 60;
  ios_base::sync_with_stdio(false);
  long long l, r, k;
  cin >> l >> r >> k;
  if (r - l + 1 <= 4LL) {
    int size = r - l + 1;
    vector<long long> nums(size);
    for (int i = 0; i < size; ++i) nums[i] = l + i;
    vector<bool> bm(size, false);
    brute(0, k, bm, nums);
    cout << mn << endl;
    int bits = 0;
    long long ans = 0;
    for (int i = 0; i < int(mnbm.size()); ++i) {
      if (mnbm[i]) {
        ++bits;
        ans ^= nums[i];
      }
    }
    cout << bits << endl;
    for (int i = 0; i < int(mnbm.size()); ++i) {
      if (mnbm[i]) {
        cout << nums[i] << " ";
      }
    }
    cout << endl;
  } else if (k == 1) {
    cout << l << endl << 1 << endl << l;
  } else if (k == 2) {
    if (l & 1) ++l;
    cout << 1 << endl << 2 << endl << l << " " << l + 1 << endl;
  } else if (k >= 4) {
    if (l & 1) ++l;
    cout << 0 << endl
         << 4 << endl
         << l << " " << l + 1 << " " << l + 2 << " " << l + 3 << endl;
  } else {
    if ((msb(l) | (msb(l) << 1)) <= r) {
      cout << 0 << endl << 3 << endl;
      long long y = l;
      long long x = msb(l);
      y &= ~x;
      y |= (x << 1);
      x |= (x << 1);
      cout << x << " " << y << " " << l << endl;
    } else {
      if (l & 1) ++l;
      cout << 1 << endl << 2 << endl << l << " " << l + 1 << endl;
    }
  }
}