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 cal(long long k, long long l, long long r) {
  if (k == 1) return l;
  if (k == 2) {
    long long ans = 2e18;
    for (long long i = l; i <= r; i++) {
      for (long long j = i + 1; j <= r; j++) {
        ans = min(ans, (i ^ j));
      }
    }
    return ans;
  }
  if (k == 3) {
    long long ans = 2e18;
    for (long long i = l; i <= r; i++) {
      for (long long j = i + 1; j <= r; j++) {
        for (long long t = j + 1; t <= r; t++) {
          ans = min(ans, (i ^ j) ^ t);
        }
      }
    }
    return ans;
  }
  return l ^ (l + 1) ^ (l + 2) ^ (l + 3);
}
int main() {
  long long l, r, k;
  cin >> l >> r >> k;
  if (r - l + 1 >= 5) {
    if (k >= 4) {
      cout << 0 << endl;
      cout << 4 << endl;
      cout << l + l % 2 << " " << l + l % 2 + 1 << " " << l + 2 + l % 2 << " "
           << l + 3 + l % 2 << endl;
      return 0;
    } else if (k == 1) {
      cout << l << endl;
      cout << 1 << endl;
      cout << l << endl;
      return 0;
    } else if (k == 2) {
      cout << 1 << endl;
      cout << 2 << endl;
      cout << l + l % 2 << " " << l + 1 + l % 2 << endl;
      return 0;
    } else if (k == 3) {
      long long x = l, num = 1;
      while (x) {
        x /= 2;
        num *= 2;
      }
      num += num / 2;
      long long v = num ^ l;
      if (num >= l && num <= r && v >= l && v <= r) {
        cout << 0 << endl;
        cout << 3 << endl;
        cout << l << " " << num << " " << v << endl;
        return 0;
      } else {
        cout << 1 << endl;
        cout << 2 << endl;
        cout << l % 2 + l << " " << l + 1 + l % 2 << endl;
        return 0;
      }
    }
  } else {
    long long ans = 2e18, x = 0;
    for (long long i = 1; i <= k; i++) {
      if (ans > cal(i, l, r)) {
        ans = cal(i, l, r);
        x = i;
      }
    }
    cout << ans << endl;
    cout << x << endl;
    if (x == 1) {
      cout << l << endl;
      return 0;
    }
    if (x == 2) {
      for (long long i = l; i <= r; i++) {
        for (long long j = i + 1; j <= r; j++) {
          if ((i ^ j) == ans) {
            cout << i << " " << j << endl;
            return 0;
          }
        }
      }
    }
    if (x == 3) {
      for (long long i = l; i <= r; i++) {
        for (long long j = i + 1; j <= r; j++) {
          for (long long t = j + 1; t <= r; t++) {
            if (((i ^ j) ^ t) == ans) {
              cout << i << " " << j << " " << t << endl;
              return 0;
            }
          }
        }
      }
    }
    if (x == 4) {
      cout << l << " " << l + 1 << " " << l + 2 << " " << l + 3 << endl;
      return 0;
    }
  }
}