Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long N = 1e5 + 5;
long long mod = 1e9 + 7;
vector<long long> adj[N];
vector<bool> vis(N);
void dfs(long long x) {
  vis[x] = true;
  for (long long i = 0; i < adj[x].size(); i++) {
    if (!vis[adj[x][i]]) dfs(adj[x][i]);
  }
}
long long d, x, y;
void extendedEuclid(long long A, long long B) {
  if (B == 0) {
    d = A;
    x = 1;
    y = 0;
  } else {
    extendedEuclid(B, A % B);
    long long temp = x;
    x = y;
    y = temp - (A / B) * y;
  }
}
long long modInverse(long long A, long long M) {
  extendedEuclid(A, M);
  return (x % M + M) % M;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long a, b;
  cin >> a >> b;
  if ((abs(a - b)) % 2 == 1 || a > b) {
    cout << -1;
    return 0;
  }
  vector<long long> a1(64, 0), b1(64, 0);
  for (long long i = 0; i < 64; i++) {
    if (a % 2 == 1) a1[i] = 1;
    a /= 2;
  }
  for (long long i = 0; i < 64; i++) {
    if (b % 2 == 1) b1[i] = 1;
    b /= 2;
  }
  long long l = 0;
  for (long long i = 63; i > 0; i--) {
    if (b1[i] % 2 != a1[i]) {
      b1[i - 1] += 2;
      b1[i]--;
    }
  }
  for (long long i = 0; i < 63; i++) {
    if (b1[i] < 0) {
      b1[i] += 4;
      b1[i + 1] -= 2;
    }
  }
  long long num = 0, c1 = 0, c2 = 0, c3 = 0, k = 1;
  for (long long i = 0; i < 64; i++) {
    if (b1[i] == 1) {
      c1 += k;
    } else if (b1[i] == 2) {
      c1 += k;
      c2 += k;
    } else if (b1[i] == 3) {
      c1 += k;
      c2 += k;
      c3 += k;
    }
    k = k << 1;
    num = max(num, b1[i]);
  }
  cout << num << "\n";
  if (c1 > 0) cout << c1 << " ";
  if (c2 > 0) cout << c2 << " ";
  if (c3 > 0) cout << c3 << " ";
  return 0;
}