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;
void _fill_int(int* p, int val, int rep) {
  int i;
  for (i = 0; i < rep; i++) p[i] = val;
}
signed long long L, R, K;
void solve() {
  int f, i, j, k, l, x, y;
  cin >> L >> R >> K;
  if (K == 1) return (void)printf("%lld\n1\n%lld\n", L, L);
  if (R - L == 1) {
    if ((L ^ R) < L)
      return (void)printf("%lld\n2\n%lld %lld\n", L ^ R, L, L + 1);
    return (void)printf("%lld\n1\n%lld\n", L, L);
  }
  if (R - L <= 10) {
    int mask;
    for (mask = 0; mask < 1 << (R - L + 1); mask++)
      if (mask) {
        signed long long tot = 0;
        for (i = 0; i < R - L + 1; i++)
          if (mask & (1 << i)) tot ^= L + i;
        if (tot == 0 && __builtin_popcount(mask) <= K) {
          (void)printf("0\n%d\n", __builtin_popcount(mask));
          for (i = 0; i < R - L + 1; i++)
            if (mask & (1 << i)) (void)printf("%lld ", L + i);
          return;
        }
      }
  }
  if (K >= 3) {
    int lb = 0;
    while (1LL << (lb + 1) <= L) lb++;
    signed long long aa = (1LL << (1 + lb)) + (L - (1LL << lb));
    signed long long bb = (1LL << (1 + lb)) + (1LL << lb);
    if (bb <= R)
      return (void)printf("%lld\n3\n%lld %lld %lld\n", 0LL, L, aa, bb);
  }
  L = (L + 1) / 2 * 2;
  if (L + 3 < R && K >= 4)
    return (void)printf("0\n4\n%lld %lld %lld %lld\n", L, L + 1, L + 2, L + 3);
  (void)printf("1\n2\n%lld %lld\n", L, L + 1);
}
int main(int argc, char** argv) {
  string s;
  if (argc == 1) ios::sync_with_stdio(false);
  for (int i = 1; i < argc; i++) s += argv[i], s += '\n';
  for (int i = s.size() - 1; i >= 0; i--) ungetc(s[i], stdin);
  solve();
  return 0;
}