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;
vector<long long> small(long long l, long long r, int k) {
  vector<long long> x;
  while (l <= r) {
    x.push_back(l);
    l++;
  }
  int n = (int)x.size();
  vector<long long> ans;
  long long best = -1;
  for (int i = 1; i < (1 << n); i++) {
    if (__builtin_popcount(i) > k) continue;
    vector<long long> cur;
    long long cc = 0;
    for (int j = 0; j < n; j++) {
      if (i & (1 << j)) {
        cur.push_back(x[j]);
        cc ^= x[j];
      }
    }
    if (best == -1 || cc < best) {
      best = cc;
      ans = cur;
    }
  }
  return ans;
}
void pt(vector<long long> &ans) {
  long long cc = 0;
  for (vector<long long>::iterator it = ans.begin(); it != ans.end(); it++) {
    cc ^= *it;
  }
  cout << cc << endl;
  cout << ans.size() << endl;
  int n = (int)ans.size();
  for (int i = 0; i < n; i++) {
    cout << ans[i];
    if (i + 1 == n)
      cout << endl;
    else
      cout << " ";
  }
}
long long getX(vector<long long> &ans) {
  long long cc = 0;
  for (vector<long long>::iterator it = ans.begin(); it != ans.end(); it++) {
    cc ^= *it;
  }
  return cc;
}
pair<long long, long long> dfs(long long l, long long r) {
  if (r - l <= 10) {
    vector<long long> c = small(max(l, 1LL), r, 3);
    sort(c.begin(), c.end());
    if (c.size() < 3)
      return pair<long long, long long>(-1, -1);
    else
      return pair<long long, long long>(c[0], c[2]);
  } else {
    pair<long long, long long> cur = dfs(l / 2, r / 2);
    long long a = cur.first, b = cur.second;
    if (a == -1) return cur;
    if (2 * a < l)
      a = 2 * a + 1;
    else
      a = 2 * a;
    if (2 * b + 1 <= r)
      b = 2 * b + 1;
    else
      b = 2 * b;
    long long c = a ^ b;
    long long x[3] = {a, b, c};
    sort(x, x + 3);
    return pair<long long, long long>(x[0], x[2]);
  }
}
int main() {
  long long l, r;
  int k;
  cin >> l >> r >> k;
  vector<long long> ans;
  if (r - l + 1 <= 10) {
    ans = small(l, r, k);
  } else if (k >= 4) {
    if (l % 2 == 1) l++;
    for (int i = 0; i < 4; i++) {
      ans.push_back(l);
      l++;
    }
  } else if (k == 1) {
    ans.push_back(l);
  } else if (k == 2) {
    if (l % 2 == 1) l++;
    ans.push_back(l);
    ans.push_back(l + 1);
  } else {
    if (l % 2 == 1) {
      ans.push_back(l + 1);
      ans.push_back(l + 2);
    } else {
      ans.push_back(l);
      ans.push_back(l + 1);
    }
    pair<long long, long long> cur = dfs(l, r);
    if (cur.first != -1) {
      vector<long long> cc;
      cc.push_back(cur.first);
      cc.push_back(cur.second);
      cc.push_back(cc[0] ^ cc[1]);
      ans = cc;
    }
  }
  pt(ans);
  return 0;
}