Given 2 integers $u$ and $v$, find the shortest array such that bitwise-xor of its elements is $u$, and the sum of its elements is $v$.
Input
The only line contains 2 integers $u$ and $v$ $(0 \le u,v \le 10^{18})$.
Output
If there’s no array that satisfies the condition, print “-1”. Otherwise:
The first line should contain one integer, $n$, representing the length of the desired array. The next line should contain $n$ positive integers, the array itself. If there are multiple possible answers, print any.
Examples
input
2 4
output
2 3 1
input
1 3
output
3 1 1 1
input
8 5
output
-1
input
0 0
output
0
Note
In the first sample, $3\oplus 1 = 2$ and $3 + 1 = 4$. There is no valid array of smaller length.
Notice that in the fourth sample the array is empty.
Solution:
#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
typedef double db;
mt19937 mrand(random_device{}());
const ll mod=1000000007;
int rnd(int x) { return mrand() % x;}
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
// head
ll u,v;
int main() {
scanf("%lld%lld",&u,&v);
if ((v-u)%2!=0) {
puts("-1");
return 0;
}
if (u>v) {
puts("-1");
return 0;
}
if (u==0&&v==0) {
puts("0");
return 0;
}
if (u==v) {
puts("1");
printf("%lld\n",u);
return 0;
}
ll p=(v-u)/2;
ll q=u^p;
if (p+q==v) {
puts("2");
printf("%lld %lld\n",p,q);
return 0;
}
puts("3");
printf("%lld %lld %lld\n",u,(v-u)/2,(v-u)/2);
}
