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, k;
vector<long long> V;
void solve() {
  if (k >= 4) {
    if (r - l + 1 == 4 && l % 2 == 1) {
      long long t = l;
      for (long long i = l + 1; i <= r; i++) t ^= i;
      if (!t) {
        for (int i = l; i <= r; i++) V.push_back(i);
        return;
      }
      for (long long i = l; i <= r; i++)
        for (long long j = l + 1; j <= r; j++)
          if ((i ^ j) >= l && (i ^ j) <= r) {
            V.push_back(i);
            V.push_back(j);
            V.push_back(i ^ j);
            return;
          }
      if (l & 1) l++;
      V.push_back(l);
      V.push_back(l + 1);
      return;
    } else {
      if (l & 1) l++;
      for (long long i = l; i < l + 4; i++) V.push_back(i);
      return;
    }
  }
  if (k == 3) {
    long long p;
    for (p = 1; p < l; p = p << 1 | 1)
      ;
    long long q = p + 1;
    q += q >> 1;
    if (q <= r) {
      V.push_back(p);
      V.push_back(q);
      V.push_back(q - 1);
      return;
    } else {
      if (l & 1) l++;
      V.push_back(l);
      V.push_back(l + 1);
      return;
    }
  }
  if (k == 1) {
    V.push_back(l);
    return;
  }
  if (k == 2 && r - l + 1 == 2) {
    if ((l ^ r) > l)
      V.push_back(l);
    else {
      V.push_back(l);
      V.push_back(r);
    }
    return;
  }
  if (l & 1) l++;
  V.push_back(l);
  V.push_back(l + 1);
}
int main() {
  scanf("%I64d%I64d%I64d", &l, &r, &k);
  solve();
  long long ans = 0;
  for (int i = 0; i < V.size(); i++) ans ^= V[i];
  printf("%I64d\n", ans);
  printf("%d\n", (int)V.size());
  printf("%I64d", V[0]);
  for (int i = 1; i < V.size(); i++) printf(" %I64d", V[i]);
  puts("");
}