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-10;
const double PI = atan(1.0) * 4;
const int inf = 2147483647 / 2;
const int DX[] = {1, 0, -1, 0}, DY[] = {0, 1, 0, -1};
template <class T>
bool chmin(T &a, const T &b) {
  return b < a ? a = b, 1 : 0;
}
template <class T>
bool chmax(T &a, const T &b) {
  return b > a ? a = b, 1 : 0;
}
template <class T>
int sgn(T first) {
  return (first > eps) - (first < -eps);
}
struct nod {
  long long v;
  vector<long long> an;
  nod() {
    v = 0;
    an.clear();
  }
  void push_back(long long first) {
    v ^= first;
    an.push_back(first);
  }
  bool operator<(const nod &t) const {
    if (!((int)(an).size())) return 0;
    if (!((int)(t.an).size())) return 1;
    return v < t.v;
  }
  void calc(long long l, int k) {
    if (l & 1) ++l;
    for (long long i = (l); i <= (long long)(l + k - 1); i++) push_back(i);
  }
  void OUT() {
    cout << v << endl;
    cout << ((int)(an).size()) << endl;
    for (long long i = (0); i <= (long long)(((int)(an).size()) - 1); i++) {
      if (i) cout << " ";
      cout << an[i];
    }
    cout << endl;
  }
};
nod ans;
long long l, r, k;
bool vis[111];
void dfs(long long p, long long s, int cnt) {
  if (p > r) {
    if (!cnt || cnt > k) return;
    nod t;
    for (long long i = (l); i <= (long long)(r); i++)
      if (vis[i - l]) t.push_back(i);
    chmin(ans, t);
    return;
  }
  dfs(p + 1, s, cnt);
  vis[p - l] = 1;
  dfs(p + 1, s ^ p, cnt + 1);
  vis[p - l] = 0;
}
long long find(long long t) {
  long long ret = 1;
  while ((ret << 1) <= t) ret <<= 1;
  return ret;
}
int main() {
  while (cin >> l >> r >> k) {
    ans.an.clear();
    if (r - l + 1 <= 5)
      memset(vis, 0, sizeof(vis)), dfs(l, 0, 0);
    else if (k == 1)
      ans.push_back(l);
    else if (k == 2 || k >= 4)
      ans.calc(l, k > 4 ? 4 : k);
    else {
      long long lp = find(l), make_pair = lp << 1, rp = find(r);
      if (make_pair == (rp << 1))
        ans.calc(l, 2);
      else if (make_pair < rp) {
        ans.push_back(l);
        ans.push_back(make_pair);
        ans.push_back(l ^ make_pair);
      } else {
        if ((rp ^ lp) > r)
          ans.calc(l, 2);
        else {
          ans.push_back(rp - 1);
          ans.push_back(rp ^ lp);
          ans.push_back((rp ^ lp) - 1);
        }
      }
    }
    ans.OUT();
  }
  return 0;
}