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;
long long l, r;
int k, cnt;
bool vis[5];
long long min_val, ans[5];
void dfs(long long cur, int used) {
  if (used > k) return;
  if (cur == r + 1) {
    if (!used) return;
    long long tmp = 0;
    for (long long i = l; i <= r; i++) {
      if (vis[i - l]) tmp ^= i;
    }
    if (tmp < min_val) {
      min_val = tmp;
      cnt = used;
      int j = 0;
      for (long long i = l; i <= r; i++) {
        if (vis[i - l]) ans[j++] = i;
      }
    }
    return;
  }
  vis[cur - l] = false;
  dfs(cur + 1, used);
  vis[cur - l] = true;
  dfs(cur + 1, used + 1);
}
int main() {
  scanf("%I64d%I64d%d", &l, &r, &k);
  min_val = r + 1;
  if (r - l <= 4)
    dfs(l, 0);
  else {
    if (k == 1) {
      min_val = l;
      cnt = 1;
      ans[0] = l;
    } else if (k == 2) {
      min_val = 1;
      cnt = 2;
      ans[0] = l + 1;
      if (l % 2)
        ans[1] = l + 2;
      else
        ans[1] = l;
    } else if (k >= 4) {
      min_val = 0;
      cnt = 4;
      ans[0] = l + 1;
      ans[1] = l + 2;
      ans[2] = l + 3;
      if (l % 2)
        ans[3] = l + 4;
      else
        ans[3] = l;
    } else {
      long long mi = 1;
      while (mi <= l) mi <<= 1;
      if (l == mi) l += 1;
      if (mi + mi / 2 <= r) {
        cnt = 3;
        min_val = 0;
        ans[0] = l + mi / 2;
        ans[1] = mi + mi / 2;
        ans[2] = l;
      } else {
        cnt = 2;
        min_val = 1;
        ans[0] = (l / 2 + 1) * 2;
        ans[1] = (l / 2 + 1) * 2 + 1;
      }
    }
  }
  printf("%I64d\n", min_val);
  printf("%d\n", cnt);
  for (int i = 0; i < cnt; i++) printf("%I64d ", ans[i]);
  return 0;
}