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 long long INF = 1e18;
vector<long long> getCandidate3(long long l, long long r) {
  vector<long long> v;
  long long p = 1;
  while (p <= l) {
    p *= 2;
  }
  if (p + p / 2 <= r) {
    long long x = p + p / 2;
    v.push_back(l);
    v.push_back(x);
    v.push_back(x ^ l);
  } else
    return v;
}
vector<long long> getCandidate2_1(long long l, long long r) {
  vector<long long> v;
  if (l + 1 <= r) {
    for (int i = 0; i < 2; i++) v.push_back(l + i);
  }
  return v;
}
vector<long long> getCandidate2_2(long long l, long long r) {
  vector<long long> v;
  if (l + 2 <= r) {
    for (int i = 1; i < 3; i++) v.push_back(l + i);
  }
  return v;
}
vector<long long> getCandidate1(long long l, long long r) {
  vector<long long> v;
  v.push_back(l);
  return v;
}
vector<long long> getCandidate4_1(long long l, long long r) {
  vector<long long> v;
  if (l + 3 <= r) {
    for (int i = 0; i < 4; i++) v.push_back(l + i);
  }
  return v;
}
vector<long long> getCandidate4_2(long long l, long long r) {
  vector<long long> v;
  if (l + 4 <= r) {
    for (int i = 1; i < 5; i++) v.push_back(l + i);
  }
  return v;
}
long long getScore(vector<long long> v) {
  if ((int)v.size() == 0) return INF;
  long long ans = 0;
  for (int i = 0; i < (int)v.size(); i++) ans ^= v[i];
  return ans;
}
long long best = INF;
vector<long long> ans;
void update(vector<long long> v) {
  long long score = getScore(v);
  if (score < best) {
    best = score;
    ans = v;
  }
}
int main() {
  ios::sync_with_stdio(false);
  long long l, r, k;
  cin >> l >> r >> k;
  if (k >= 1) update(getCandidate1(l, r));
  if (k >= 2) {
    update(getCandidate2_1(l, r));
    update(getCandidate2_2(l, r));
  }
  if (k >= 3) update(getCandidate3(l, r));
  if (k >= 4) {
    update(getCandidate4_1(l, r));
    update(getCandidate4_2(l, r));
  }
  cout << best << '\n';
  cout << ans.size() << '\n';
  for (int i = 0; i < (int)ans.size(); i++) cout << ans[i] << " ";
  cout << '\n';
  return 0;
}