暴力+构造

If r - l ≤ 4 we can all subsets of size not greater than k. Else, if k = 1, obviously that answer is l. If k = 2, answer is 1, because xor of numbers 2x and 2x + 1 equls 1. If k ≥ 4 answer is 0 because xor of to pairs with xor 1 is 0.

If k = 3, we can choose numbers 2x and 2x + 1 with xor 1. So we need to know, if we can get xor equals 0. Suppose that there are 3 such numbers x, y and z (r ≥ x > y > z ≥ l) with xor equals 0. Consider the most non-zero bit of numberx. At the same bit of y it's also 1, because xor equls 0, and y > z. Consider the next bit of numbers. If z have 0 there, we have to do next: set the previous bit of numbers x and y equals 0, and set current bit equals 1. Obviously xor still equals 0, z hadn't changed and numbers x and y stood closer to z, so they are still at [l, r].And x > y.Consider the next bit of numbers. If z has zero here than we will change most bits of x и y at the same way and so on. z > 0, so somewhen we will get to bit in which z has 1. Since xorequals 0, the same bit of x would be 1 because x > y, and y would have 0 there. At the next bits we will change bit in x to 0, and in numbers y and z to 1.Finally z would be greater or equal than before, and x would be less or greater than before, and x > y > z would be correct. So, we have the next: if such numbers x, y and z exist than also exist numbers:

1100…000

1011…111

0111…111

with xor equals 0. There are not much such triples, so we can check them.

D. Little Victor and Set

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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  the following inequality holds l ≤ x ≤ r;
• 1 ≤ |S| ≤ k;
• lets denote the i-th element of the set S as si; value  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.

Sample test(s)

input

8 15 3

output

1210 11

input

8 30 7

output

0514 9 28 11 16

Note

Operation  represents the operation of bitwise exclusive OR. In other words, it is the XOR operation.

#include 
     
      #include 
      
       #include 
       
        #include 
        
         using namespace std;typedef long long int LL;LL L,R,K;LL ans=0x3f3f3f3f3f3f3f3f;int main(){    cin>>L>>R>>K;    if(R-L+1<=4)    {        LL m=R-L+1;        LL sig=0;        for(LL i=1;i<(1LL<
         
          K) continue;            for(LL j=0;j
          
           =L) { flag=true; cout<<0<
           
            <<3<