Given 2 integers u and v, find the shortest array such that [bitwise-xor](https://en.wikipedia.org/wiki/Bitwise_operation#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 ≤ u,v ≤ 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⊕ 1 = 2 and 3 + 1 = 4. There is no valid array of smaller length.

Notice that in the fourth sample the array is empty.
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
const int M = 1e6 + 5;
const int MOD = 1e9 + 7;
const int CM = 998244353;
const int INF = 0x3f3f3f3f;
const double eps = 1e-6;
template <typename T>
T MAX(T a, T b) {
  return a > b ? a : b;
}
template <typename T>
T MIN(T a, T b) {
  return a > b ? b : a;
}
template <typename T>
T GCD(T a, T b) {
  return b == 0 ? a : GCD(b, a % b);
}
template <typename T>
T LCM(T a, T b) {
  return a / GCD(a, b) * b;
}
template <typename T>
T ABS(T a, T b) {
  return a > 0 ? a : -a;
}
template <typename T>
T ADD(T a, T b, T MOD) {
  return (a + b) % MOD;
}
template <typename T>
T DEL(T a, T b, T MOD) {
  return ((a - b) % MOD + MOD) % MOD;
}
void solve(int kase) {
  long long u, v;
  scanf("%lld%lld", &u, &v);
  if (u > v || u % 2 != v % 2)
    printf("-1\n");
  else if (u == v) {
    if (u == 0)
      printf("0\n");
    else
      printf("1\n%lld\n", u);
  } else {
    long long a = (v - u) / 2;
    if ((a ^ u) == a + u) {
      printf("2\n%lld %lld\n", (u + v) / 2, a);
    } else {
      printf("3\n%lld %lld %lld", u, a, a);
    }
  }
}
const bool DUO = 0;
void TIME() {
  clock_t start, finish;
  double totaltime;
  start = clock();
  if (DUO) {
    int Kase = 0;
    cin >> Kase;
    for (int kase = 1; kase <= Kase; kase++) solve(kase);
  } else
    solve(1);
  finish = clock();
  totaltime = (double)(finish - start) / CLOCKS_PER_SEC;
  printf("\nTime:%lfms\n", totaltime * 1000);
}
int main() {
  if (DUO) {
    int Kase = 0;
    cin >> Kase;
    for (int kase = 1; kase <= Kase; kase++) solve(kase);
  } else
    solve(1);
  return 0;
}
