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-8;
const long long MOD = 1000000007;
const long long INF = 0x3f3f3f3f;
long long l, r, k;
vector<long long> g;
long long ans;
int main() {
  long long i, j, u, v, w;
  while (cin >> l >> r >> k) {
    ans = l;
    g.clear();
    g.push_back(l);
    if (r - l + 1 > 5) {
      if (k >= 1) {
        ans = l;
        g.clear();
        g.push_back(l);
      }
      if (k >= 2) {
        ans = 1;
        g.clear();
        if (l % 2 == 0) {
          g.push_back(l);
          g.push_back(l + 1);
        } else {
          g.push_back(l + 1);
          g.push_back(l + 2);
        }
      }
      if (k >= 3) {
        int flag = 0;
        for (i = 0; (3LL << i) <= r; i++) {
          if ((1LL << i) <= l && l < (2LL << i)) {
            g.clear();
            ans = 0;
            g.push_back(3LL << i);
            g.push_back(l ^ (3LL << i));
            g.push_back(l);
            ans = 0;
            flag = 1;
            break;
          }
        }
      }
      if (k >= 4) {
        ans = 0;
        g.clear();
        if (l % 2 == 0) {
          for (i = l, j = 0; j < 4; i++, j++) g.push_back(i);
        } else {
          for (i = l + 1, j = 0; j < 4; i++, j++) g.push_back(i);
        }
      }
    } else {
      u = r - l + 1;
      for (i = 1; i < (1LL << u); i++) {
        for (j = 0, v = 0, w = 0; j < u; j++) {
          if (i & (1 << j)) {
            v++;
            w ^= l + j;
          }
        }
        if (v <= k) {
          if (w < ans) {
            g.clear();
            ans = w;
            for (j = 0; j < u; j++) {
              if (i & (1LL << j)) g.push_back(l + j);
            }
          }
        }
      }
    }
    cout << ans << endl;
    cout << g.size() << endl;
    for (i = 0; i < g.size(); i++) cout << g[i] << " ";
    cout << endl;
  }
  return 0;
}