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 l, r, k;
vector<long long> v;
long long b[10], bn;
long long c[10], cn, miny;
void dfs(long long x, long long y, int cnt) {
  if (cnt > k) return;
  if (x > r) {
    if (cnt != 0 && y < miny) {
      cn = bn;
      int i;
      for (i = 0; i < (bn); i++) c[i] = b[i];
      miny = y;
    }
    return;
  }
  b[bn++] = x;
  dfs(x + 1, y ^ x, cnt + 1);
  bn--;
  dfs(x + 1, y, cnt);
}
long long farm() {
  long long i;
  v.clear();
  if (k == 1) {
    v.push_back(l);
    return l;
  }
  if (k >= 4) {
    long long wa = (long long)(l / 4) * 4;
    while (wa < l) wa += 4;
    if (wa + 3 <= r) {
      v.push_back(wa);
      v.push_back(wa + 1);
      v.push_back(wa + 2);
      v.push_back(wa + 3);
      return 0;
    }
    bn = 0;
    miny = 1LL << 60;
    dfs(l, 0, 0);
    for (i = 0; i < (cn); i++) v.push_back(c[i]);
    return miny;
  }
  if (k == 3) {
    int lwei, rwei;
    lwei = 0;
    rwei = 0;
    long long t = l;
    while (t) t >>= 1, lwei++;
    t = r;
    while (t) t >>= 1, rwei++;
    if (rwei >= 2 && lwei < rwei) {
      long long q[3];
      for (i = lwei + 1 - 2; i <= lwei + 2 - 2; i++) {
        q[0] = (3LL << i);
        q[1] = q[0] - 1;
        q[2] = (1LL << (i + 1)) - 1;
        bool dou = 0;
        for (int j = 0; j < 3; j++)
          if (q[j] < l || q[j] > r) {
            dou = 1;
            break;
          }
        if (dou) continue;
        v.push_back(q[0]);
        v.push_back(q[1]);
        v.push_back(q[2]);
        return 0;
      }
    }
  }
  if (l == 1) {
    v.push_back(1);
    return 1;
  }
  long long wa = ((long long)((l + 1) / 2)) * 2;
  if (wa >= l && wa + 1 <= r) {
    v.push_back(wa);
    v.push_back(wa + 1);
    return 1;
  }
  if (l < (l ^ r)) {
    v.push_back(l);
    return l;
  } else {
    v.push_back(l);
    v.push_back(r);
    return l ^ r;
  }
}
int main() {
  long long i;
  scanf("%I64d%I64d%I64d", &l, &r, &k);
  long long ans = farm();
  printf("%I64d\n", ans);
  long long maxi = v.size();
  printf("%I64d\n", maxi);
  if (maxi > 0) printf("%I64d", v[0]);
  for (i = 1; i < maxi; i++) printf(" %I64d", v[i]);
  return 0;
}