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 double EPS = 1e-8;
const long long SZ = 200010, SSZ = 100010, APB = 26, one = 93,
                INF = 0x7fffffffffffffff, mod = 1000000007;
const int intINF = 0x3f3f3f3f;
long long ll, rr, n;
int tim;
void calc() {
  long long minv = INF;
  vector<long long> res;
  for (long long i = 1LL; i < (1LL << (rr - ll + 1LL)); ++i) {
    vector<long long> cur;
    long long tmp = 0;
    if (__builtin_popcount(i) <= n) {
      for (long long j = 0; j < (rr - ll + 1LL); ++j) {
        if (i & (1LL << j)) {
          cur.push_back(ll + j);
          tmp ^= ll + j;
        }
      }
      if (tmp < minv) {
        minv = tmp;
        res = cur;
      }
    }
  }
  cout << minv << endl;
  cout << res.size() << endl;
  for (long long i = 0; i < res.size(); ++i) {
    cout << res[i] << " ";
  }
}
bool chk() {
  long long num = 0;
  for (;; ++num) {
    if ((1LL << num) - 1LL >= ll) break;
  }
  long long z = (1LL << num) - 1LL;
  long long x = (1LL << num) + (1LL << (num - 1LL));
  long long y = x - 1LL;
  if (x <= rr) {
    cout << 0 << endl;
    cout << 3 << endl;
    cout << x << " " << y << " " << z << endl;
    return 1LL;
  } else
    return 0;
}
void init() {
  cin >> ll >> rr >> n;
  if (rr - ll <= 6)
    calc();
  else if (n == 1LL)
    cout << ll << endl << 1LL << endl << ll << endl;
  else if (n == 2) {
    cout << 1LL << endl << 2 << endl;
    if (ll & 1LL) {
      cout << ll + 1LL << " " << ll + 2 << endl;
    } else
      cout << ll << " " << ll + 1LL << endl;
  } else if (n == 3) {
    if (chk()) {
    } else {
      cout << 1LL << endl;
      cout << 2LL << endl;
      if (ll & 1LL) {
        cout << ll + 1LL << " " << ll + 2 << endl;
      } else
        cout << ll << " " << ll + 1LL << endl;
    }
  } else {
    cout << 0LL << endl << 4 << endl;
    if (ll & 1LL)
      cout << ll + 1LL << " " << ll + 2 << " " << ll + 3 << " " << ll + 4
           << endl;
    else
      cout << ll << " " << ll + 1LL << " " << ll + 2 << " " << ll + 3 << endl;
  }
}
void work() {}
int main() {
  std::ios::sync_with_stdio(0);
  long long casenum;
  {
    init();
    work();
  }
  return 0;
}