题目

思路：

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define pb push_back
#define fi first
#define se second
#define lson p << 1
#define rson p << 1 | 1
const int maxn = 1e6 + 5, inf = 1e18, maxm = 4e4 + 5;
const int N = sqrt(1e9) + 1;
 const int mod = 1e9 + 7;
// const int mod = 998244353;
//const __int128 mod = 212370440130137957LL;
// int a[505][5005];
// bool vis[505][505];

int a[maxn], b[maxn];
bool vis[maxn];
string s[maxn];
int n, m;

struct Node{
    int val, id;
    bool operator<(const Node &u)const{
        return val < u.val;
    }
};
// Node c[maxn];
int ans[maxn];
int pre[maxn];

int f[20][10][10];
//long long ? maxn ? n? m?

void solve(){
    int res = 0;
    int q, k;
    int x;
    cin >> n >> m;
    // for(int i = 1; i <= n; i++){
    //     cin >> a[i];
    // }
    if((n ^ m) < n){//m的最高有效位在a或b都只需要一次操作
        cout << 1 << '\n';
        cout << n << ' ' << m << '\n';
        return;
    }
    if((n & (n - 1)) == 0){//n为2的幂次
        cout << -1 << '\n';
        return;
    }
    int a = -1, b = -1;
    int c = -1;
    for(int j = 62; j >= 0; j--){
        if(n >> j & 1){
            if(a == -1){
                a = j;
            }
            else if(b == -1){
                b = j;
            }
        }
        if(m >> j & 1){
            if(c == -1){
                c = j;
            }
        }
    }
    if(c > b && c < a){
        cout << -1 << '\n';
        return;
    }
    cout << 2 << '\n';
    //第一步：先忽略n的最高有效位a，把其余位变成和m一样，第二步：消去最高有效位得到m
    cout << n << ' ' << ((n - (1LL << a)) ^ m) << ' ' << m << '\n';
}   
    
signed main(){
    ios::sync_with_stdio(0);
    cin.tie(0);

    int T = 1;
    cin >> T;
    while (T--)
    {
        solve();
    }
    return 0;
}