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;
template <class T>
int getbit(T s, int i) {
  return (s >> i) & 1;
}
template <class T>
T onbit(T s, int i) {
  return s | (T(1) << i);
}
template <class T>
T offbit(T s, int i) {
  return s & (~(T(1) << i));
}
template <class T>
int cntbit(T s) {
  return __builtin_popcount(s);
}
template <class T>
T gcd(T a, T b) {
  T r;
  while (b != 0) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
template <class T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
long long l, r, k;
void cal(long long x) {
  long long d = ((x + 1) / 2) * 2;
  cout << 1 << endl;
  cout << 2 << endl;
  cout << d << " " << d + 1 << endl;
}
long long cal(long long x, long long y) {
  int M = 0;
  for (int i = 0; i < (60); ++i)
    if (getbit(x, i) && !getbit(y, i)) {
      M = i;
    }
  return (1ll << (M + 1)) - 1;
}
void solve() {
  long long MAX = 1000000005 * 1ll * 100000;
  if (r - l <= 7) {
    vector<long long> RR;
    long long a[10];
    long long res = MAX;
    long long d = r - l + 1;
    for (int i = 0; i < (d); ++i) a[i] = l + i;
    do {
      for (int i = (1); i <= (min(k, d)); ++i) {
        long long ret = 0;
        for (int j = 0; j < (i); ++j) ret ^= a[j];
        if (ret < res) {
          res = ret;
          RR.clear();
          for (int j = 0; j < (i); ++j) RR.push_back(a[j]);
        }
      }
    } while (next_permutation(a, a + d));
    cout << res << endl;
    cout << ((int)(RR).size()) << endl;
    for (int i = 0; i < (((int)(RR).size())); ++i) cout << RR[i] << " ";
    return;
  } else if (k >= 4) {
    long long d = ((l + 3) / 4) * 4;
    cout << 0 << endl;
    cout << 4 << endl;
    for (int i = 0; i < (4); ++i) cout << d + i << " ";
    return;
  } else {
    long long t = 1, T;
    while (t + t <= r) t += t;
    T = t / 2;
    if (t < l) {
      cal(l);
    } else {
      long long dd = cal(r, t);
      if (dd >= l) {
        cout << 0 << endl;
        cout << 3 << endl;
        cout << dd << " " << (dd ^ r) << " " << r << endl;
      } else if (T > l) {
        cout << 0 << endl;
        cout << 3 << endl;
        cout << T - 1 << " " << T << " " << t - 1 << endl;
      } else {
        cal(l);
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> l >> r >> k;
  if (k == 1) {
    cout << l << endl;
    cout << 1 << endl;
    cout << l << endl;
    return 0;
  } else if (k == 2) {
    if (r - l == 1) {
      if (l < (r ^ l)) {
        cout << l << endl;
        cout << 1 << endl;
        cout << l << endl;
      } else {
        cout << (l ^ r) << endl;
        cout << 2 << endl;
        cout << l << " " << r << endl;
      }
    } else {
      cal(l);
    }
    return 0;
  } else
    solve();
  return 0;
}