#include <bits/stdc++.h>
using namespace std;
long long int visited[2000005];
long long int a[2000005], c[2000005], ans = 0;
vector<vector<int>> adj;
vector<int> dp;
int n, m;
void addEdge(vector<long long int> adj[], long long int a, long long int b) {
  adj[a - 1].push_back(b - 1);
  adj[b - 1].push_back(a - 1);
}
void dfs(vector<long long int> adj[], long long int x) {
  visited[x] = true;
  for (auto it : adj[x]) {
    if (!visited[it]) dfs(adj, it);
  }
}
void dfs1(vector<long long int> adj[], long long int i) {
  visited[i] = 1;
  long long int x = a[i];
  if (!visited[x])
    dfs1(adj, x);
  else if (visited[x] == 1) {
    long long int p = x, mn = c[i];
    while (p != i) {
      mn = min(mn, c[p]);
      p = a[p];
    }
    ans += mn;
  }
  visited[i] = 2;
  return;
}
void bfs(int s) {
  dp = vector<int>(n, 1e9);
  dp[s] = 0;
  queue<int> q;
  q.push(s);
  while (!q.empty()) {
    int v = q.front();
    q.pop();
    for (auto to : adj[v]) {
      if (dp[to] == 1e9) {
        dp[to] = dp[v] + 1;
        q.push(to);
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  unsigned long long int a, b;
  cin >> a >> b;
  if (a < b || a % 2 != b % 2)
    cout << -1;
  else {
    unsigned long long int x, y;
    x = (a - b) / 2;
    y = x + b;
    if (x ^ y == b && x + y == a)
      cout << x << " " << y;
    else
      cout << -1;
  }
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

Bitwise exclusive OR (or bitwise addition modulo two) is a binary operation which is equivalent to applying logical exclusive OR to every pair of bits located on the same positions in binary notation of operands. In other words, a binary digit of the result is equal to 1 if and only if bits on the respective positions in the operands are different.

For example, if X = 10910 = 11011012, Y = 4110 = 1010012, then: 

X xor Y = 6810 = 10001002. 

Write a program, which takes two non-negative integers A and B as an input and finds two non-negative integers X and Y, which satisfy the following conditions: 

  * A = X + Y
  * B = X xor Y, where xor is bitwise exclusive or. 
  * X is the smallest number among all numbers for which the first two conditions are true. 

Input

The first line contains integer number A and the second line contains integer number B (0 ≤ A, B ≤ 264 - 1).

Output

The only output line should contain two integer non-negative numbers X and Y. Print the only number -1 if there is no answer.

Examples

Input

142
76


Output

33 109

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long int visited[2000005];
long long int a[2000005], c[2000005], ans = 0;
vector<vector<int>> adj;
vector<int> dp;
int n, m;
void addEdge(vector<long long int> adj[], long long int a, long long int b) {
  adj[a - 1].push_back(b - 1);
  adj[b - 1].push_back(a - 1);
}
void dfs(vector<long long int> adj[], long long int x) {
  visited[x] = true;
  for (auto it : adj[x]) {
    if (!visited[it]) dfs(adj, it);
  }
}
void dfs1(vector<long long int> adj[], long long int i) {
  visited[i] = 1;
  long long int x = a[i];
  if (!visited[x])
    dfs1(adj, x);
  else if (visited[x] == 1) {
    long long int p = x, mn = c[i];
    while (p != i) {
      mn = min(mn, c[p]);
      p = a[p];
    }
    ans += mn;
  }
  visited[i] = 2;
  return;
}
void bfs(int s) {
  dp = vector<int>(n, 1e9);
  dp[s] = 0;
  queue<int> q;
  q.push(s);
  while (!q.empty()) {
    int v = q.front();
    q.pop();
    for (auto to : adj[v]) {
      if (dp[to] == 1e9) {
        dp[to] = dp[v] + 1;
        q.push(to);
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  unsigned long long int a, b;
  cin >> a >> b;
  if (a < b || a % 2 != b % 2)
    cout << -1;
  else {
    unsigned long long int x, y;
    x = (a - b) / 2;
    y = x + b;
    if (x ^ y == b && x + y == a)
      cout << x << " " << y;
    else
      cout << -1;
  }
  return 0;
}

```