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 int L, R, K;
long long int ans = 0x3f3f3f3f3f3f3f3f;
int main() {
  cin >> L >> R >> K;
  if (R - L + 1 <= 4) {
    long long int m = R - L + 1;
    long long int sig = 0;
    for (long long int i = 1; i < (1LL << m); i++) {
      long long int temp = 0;
      long long int wei = 0;
      long long int si = i;
      while (si) {
        wei++;
        si = si & (si - 1);
      }
      if (wei > K) continue;
      for (long long int j = 0; j < m; j++) {
        if (i & (1LL << j)) {
          temp ^= L + j;
        }
      }
      if (temp < ans) {
        ans = temp;
        sig = i;
      }
    }
    cout << ans << endl;
    long long int wei = 0;
    long long int tsig = sig;
    while (tsig) {
      wei++;
      tsig = tsig & (tsig - 1);
    }
    cout << wei << endl;
    for (long long int i = 0; i < m; i++) {
      if (sig & (1 << i)) {
        cout << L + i << " ";
      }
    }
    cout << endl;
  } else {
    if (K == 1) {
      cout << L << endl;
      cout << 1 << endl;
      cout << L << endl;
    } else if (K == 2) {
      if (L % 2) L++;
      cout << 1 << endl;
      cout << 2 << endl;
      cout << L << " " << L + 1 << endl;
    } else if (K == 3) {
      bool flag = false;
      long long int mx = 3, mi = 1;
      while (mx <= R) {
        if (mi >= L) {
          flag = true;
          cout << 0 << endl << 3 << endl;
          cout << mx << " " << mx - 1 << " " << mi << endl;
          break;
        }
        mx <<= 1LL;
        mi <<= 1LL;
        mi++;
      }
      if (flag == false) {
        if (L % 2) L++;
        cout << 1 << endl;
        cout << 2 << endl;
        cout << L << " " << L + 1 << endl;
      }
    } else {
      cout << 0 << endl;
      cout << 4 << endl;
      if (L % 2) L++;
      cout << L << " " << L + 1 << " " << L + 2 << " " << L + 3 << endl;
    }
  }
  return 0;
}