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;
void gao(long long x, long long &a, long long &b) {
  int cnt = 0;
  assert(__builtin_popcountll(x) >= 2);
  for (int j = 61; j >= 0; --j) {
    if ((1LL << j) & x) {
      if (cnt > 1) {
        b |= 1LL << j;
      } else if (cnt == 0) {
        a = 1LL << j;
      } else {
        b = 1LL << j;
      }
      ++cnt;
    } else if (cnt > 1) {
      a |= 1LL << j;
      b |= 1LL << j;
    }
  }
  assert(a ^ b == x);
}
int main() {
  long long l, r, k;
  cin >> l >> r >> k;
  if (k == 1) {
    cout << l << endl;
    cout << 1 << endl;
    cout << l << endl;
  } else if (r - l <= 10) {
    long long ans = -1;
    vector<long long> s;
    int n = r - l + 1;
    for (int st = 1; st < (1 << n); ++st) {
      if (__builtin_popcount(st) > k) continue;
      long long tmp = 0;
      vector<long long> curr;
      for (int i = 0; i < int(n); i++) {
        if (st & (1 << i)) {
          tmp ^= l + i;
          curr.push_back(l + i);
        }
      }
      if (ans == -1 || tmp < ans) {
        ans = tmp;
        s = curr;
      }
    }
    cout << ans << endl;
    cout << s.size() << endl;
    for (auto x : s) cout << x << ' ';
    cout << endl;
  } else if (k == 2) {
    while (l % 2 != 0) ++l;
    cout << 1 << endl;
    cout << 2 << endl;
    cout << l << ' ' << l + 1 << endl;
  } else if (k >= 4) {
    while (l % 4 != 0) ++l;
    cout << 0 << endl;
    cout << 4 << endl;
    for (int i = 0; i < int(4); i++) cout << i + l << ' ';
    cout << endl;
  } else {
    assert(k == 3);
    if (__builtin_popcountll(r) == 1) --r;
    bool found = 0;
    long long a, b;
    gao(r, a, b);
    if (b >= l)
      found = 1;
    else {
      for (int j = 61; j >= 0; --j)
        if ((1LL << j) & r) {
          r = (1LL << j) - 1;
          break;
        }
      gao(r, a, b);
      if (b >= l) found = 1;
    }
    if (!found) {
      while (l % 2 != 0) ++l;
      cout << 1 << endl;
      cout << 2 << endl;
      cout << l << ' ' << l + 1 << endl;
    } else {
      cout << 0 << endl;
      cout << 3 << endl;
      cout << r << ' ' << a << ' ' << b << endl;
    }
  }
}