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 long long inf = 1e18;
long long l, r, k, tmp, ans;
void dfs(long long i, long long sum, int now, int cnt) {
  if (cnt > k) return;
  if (i == r + 1) {
    if (ans > sum && cnt != 0) {
      ans = sum;
      tmp = now;
    }
    return;
  }
  dfs(i + 1, sum, now, cnt);
  dfs(i + 1, sum ^ i, now | (1 << (i - l)), cnt + 1);
}
int main() {
  scanf("%I64d%I64d%I64d", &l, &r, &k);
  if (r - l + 1 <= 10) {
    int count = 0, t;
    ans = inf;
    dfs(l, 0, 0, 0);
    t = tmp;
    printf("%I64d\n", ans);
    while (t != 0) {
      if (t & 1) ++count;
      t >>= 1;
    }
    printf("%d\n", count);
    count = 0;
    while (tmp != 0) {
      if (tmp & 1) printf("%I64d ", l + count);
      tmp >>= 1;
      ++count;
    }
    printf("\n");
  } else {
    if (k == 1) printf("%I64d\n1\n%I64d\n", l, l);
    if (k == 2) {
      if (l & 1) l++;
      printf("1\n2\n%I64d %I64d\n", l, l + 1);
    }
    if (k == 3) {
      long long x, y, z, m, u;
      x = 1;
      m = 0;
      y = 1;
      u = 1;
      while (u <= l) {
        u *= 2;
        ++m;
      }
      x = u + u / 2;
      y = u - 1;
      z = u + (u / 2 - 1);
      if (x <= r)
        printf("0\n3\n%I64d %I64d %I64d\n", y, z, x);
      else {
        if (l & 1) ++l;
        printf("1\n2\n%I64d %I64d\n", l, l + 1);
      }
    }
    if (k >= 4) {
      if (l & 1) ++l;
      printf("0\n4\n%I64d %I64d %I64d %I64d\n", l, l + 1, l + 2, l + 3);
    }
  }
  return 0;
}