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 = (1LL << 31) - 1;
const long long int LINF = (1LL << 63) - 1;
const int NMAX = 100000 + 5;
const int MMAX = 100000 + 5;
const int KMAX = 100000 + 5;
const int PMAX = 100000 + 5;
const int LMAX = 100000 + 5;
const int VMAX = 100000 + 5;
long long int L, R, K, S;
vector<long long int> sol;
int main() {
  scanf("%I64d%I64d%I64d", &L, &R, &K);
  if (K > 4 || (K == 4 && (L % 2 == 0 || R - L + 1 > 4))) {
    if (L % 2 == 0) {
      sol.push_back(L);
      sol.push_back(L + 1);
      sol.push_back(L + 2);
      sol.push_back(L + 3);
    } else {
      sol.push_back(L + 1);
      sol.push_back(L + 2);
      sol.push_back(L + 3);
      sol.push_back(L + 4);
    }
  } else if (K == 1) {
    sol.push_back(L);
  } else if (K == 2) {
    if (L % 2 == 0) {
      sol.push_back(L);
      sol.push_back(L + 1);
    } else if (R - L + 1 >= 3) {
      sol.push_back(L + 1);
      sol.push_back(L + 2);
    } else {
      auto x = L ^ R;
      if (x < L)
        sol.push_back(L), sol.push_back(R);
      else
        sol.push_back(L);
    }
  } else if (K >= 3) {
    long long int i, a, b, c, l = 0;
    for (i = 0; i <= 60; i++) {
      a = 2 * (1LL << i) - 1;
      b = 3 * (1LL << i) - 1;
      c = 3 * (1LL << i);
      if (L <= a && a <= R && L <= b && b <= R && L <= c && c <= R) {
        sol.push_back(a);
        sol.push_back(b);
        sol.push_back(c);
        l = 1;
        break;
      }
    }
    if (l == 0) {
      if (L % 2 == 0)
        sol.push_back(L), sol.push_back(L + 1);
      else
        sol.push_back(L + 1), sol.push_back(L + 2);
    }
  }
  S = 0;
  for (auto x : sol) S ^= x;
  printf("%I64d\n%d\n", S, sol.size());
  for (auto x : sol) printf("%I64d ", x);
  return 0;
}