The Child and Set

B. The Child and Set

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

At the children's day, the child came to Picks's house, and messed his house up. Picks was angry at him. A lot of important things were lost, in particular the favorite set of Picks.

Fortunately, Picks remembers something about his set S:

• its elements were distinct integers from 1 to limit;
• the value of  was equal to sum; here lowbit(x) equals 2k where k is the position of the first one in the binary representation of x. For example, lowbit(100102) = 102, lowbit(100012) = 12, lowbit(100002) = 100002 (binary representation).

Can you help Picks and find any set S, that satisfies all the above conditions?

Input

The first line contains two integers: sum, limit (1 ≤ sum, limit ≤ 105).

Output

In the first line print an integer n (1 ≤ n ≤ 105), denoting the size of S. Then print the elements of set S in any order. If there are multiple answers, print any of them.

If it's impossible to find a suitable set, print -1.

Examples

input

5 5

output

24 5

input

4 3

output

32 3 1

input

5 1

output

-1

Note

In sample test 1: lowbit(4) = 4, lowbit(5) = 1, 4 + 1 = 5.

In sample test 2: lowbit(1) = 1, lowbit(2) = 2, lowbit(3) = 1, 1 + 2 + 1 = 4.

题目大意：给在[1,limit]得到一些数使得他们的二进制，从高位到低位第一个 1 出

 

现的位置的和等于 sum

 

数据范围：sum, limit (1 ≤ sum, limit ≤ 100000)

 

解题思路：考虑到数目不是太大的情况下，先是将[1,limit]范围内的每个数求一下转换为二进制的时候第一个 1 的位置，然后再遍历一遍他们相加是否为 sum，之后比较综合是否为 sum，是则输出个数和具体数字，不是则输出-1

 

在这之中，有一个新的方法更方便的去转换 bitset 类去取二进制在代码中先是添加头文件#include <bitset>

 

然后在代码中的体现是这个 bitset<18> t; //声明一个含有 18 位的 t 二进制数组

 t = i;//将循环的每个数 i 传递给 t 

string str;//string 类

str = t.to_string();//bitset 类成员函数 to_string()，作用是将 t 转化成一个二进制的字符串

#include<iostream>  #include<cstring>  #include<cstdio>  #include<climits>  #include<cmath> #include <bitset>#include<algorithm>  using namespace std;int sum, limit, i, ans, k,x,n,j;int a[100008];int c[100008];int main(){cin >> sum >> limit;ans = 0;k = 0;memset(a, 0, sizeof(a));memset(c, 0, sizeof(c));for (i = limit; i >= 1; i--){bitset<18> t;t = i;string str;//string类str = t.to_string();//bitset类成员函数to_string()for (j = 1; j <= 18; j++){if (str[j] == '1'){break;}}x = j;for (j = 17; j >= x; j--){if (str[j] == '1'){c[i] = pow(2, 17-x+1-(j-x+1));break;}}}for (i = limit; i >= 1; i--){ans += c[i];if (ans > sum){ans -= c[i];}else if (ans == sum){a[i] = 1;k++;break;}else{a[i] = 1;k++;}}j = 0;if (ans == sum){cout << k << endl;for (i = 1; i <= limit; i++){if (a[i] == 1){j++;if (j == k){cout << i << endl;break;}else{cout << i << " ";}}}}else{cout << -1 << endl;}}