#include <bits/stdc++.h>
using namespace std;
int dir[4][2] = {{1, 0}, {0, 1}, {0, -1}, {-1, 0}};
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int inf = 1e9 + 5;
const double pi = 3.14159265358979323846;
const double eps = 1e-6;
const int mod = 1e9 + 7;
const int N = 2e5 + 5;
long long gcd(long long m, long long n) { return n == 0 ? m : gcd(n, m % n); }
long long lcm(long long m, long long n) { return m * n / gcd(m, n); }
bool prime(int x) {
  if (x < 2) return false;
  for (int i = 2; i * i <= x; ++i) {
    if (x % i == 0) return false;
  }
  return true;
}
long long qpow(long long m, long long k, long long mod) {
  long long res = 1, t = m;
  while (k) {
    if (k & 1) res = res * t % mod;
    t = t * t % mod;
    k >>= 1;
  }
  return res;
}
long long inv[N], f[N];
void get(int n) {
  f[0] = 1;
  for (int i = 1; i <= n; i++) f[i] = f[i - 1] * i % mod;
  inv[1] = 1;
  for (int i = 2; i <= n; i++) inv[i] = (mod - mod / i) * inv[mod % i] % mod;
  inv[0] = 1;
  for (int i = 1; i <= n; i++) inv[i] = inv[i] * inv[i - 1] % mod;
}
inline long long C(int n, int m) {
  return f[n] * inv[m] % mod * inv[n - m] % mod;
}
int main() {
  cin.tie(0);
  cout.tie(0);
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    long long l, r;
    cin >> l >> r;
    if (r >= l * 2) {
      cout << l << ' ' << 2 * l << endl;
    } else
      cout << -1 << ' ' << -1 << endl;
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

Let LCM(x, y) be the minimum positive integer that is divisible by both x and y. For example, LCM(13, 37) = 481, LCM(9, 6) = 18.

You are given two integers l and r. Find two integers x and y such that l ≤ x < y ≤ r and l ≤ LCM(x, y) ≤ r.

Input

The first line contains one integer t (1 ≤ t ≤ 10000) — the number of test cases.

Each test case is represented by one line containing two integers l and r (1 ≤ l < r ≤ 10^9).

Output

For each test case, print two integers:

  * if it is impossible to find integers x and y meeting the constraints in the statement, print two integers equal to -1; 
  * otherwise, print the values of x and y (if there are multiple valid answers, you may print any of them). 

Example

Input


4
1 1337
13 69
2 4
88 89


Output


6 7
14 21
2 4
-1 -1

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int dir[4][2] = {{1, 0}, {0, 1}, {0, -1}, {-1, 0}};
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int inf = 1e9 + 5;
const double pi = 3.14159265358979323846;
const double eps = 1e-6;
const int mod = 1e9 + 7;
const int N = 2e5 + 5;
long long gcd(long long m, long long n) { return n == 0 ? m : gcd(n, m % n); }
long long lcm(long long m, long long n) { return m * n / gcd(m, n); }
bool prime(int x) {
  if (x < 2) return false;
  for (int i = 2; i * i <= x; ++i) {
    if (x % i == 0) return false;
  }
  return true;
}
long long qpow(long long m, long long k, long long mod) {
  long long res = 1, t = m;
  while (k) {
    if (k & 1) res = res * t % mod;
    t = t * t % mod;
    k >>= 1;
  }
  return res;
}
long long inv[N], f[N];
void get(int n) {
  f[0] = 1;
  for (int i = 1; i <= n; i++) f[i] = f[i - 1] * i % mod;
  inv[1] = 1;
  for (int i = 2; i <= n; i++) inv[i] = (mod - mod / i) * inv[mod % i] % mod;
  inv[0] = 1;
  for (int i = 1; i <= n; i++) inv[i] = inv[i] * inv[i - 1] % mod;
}
inline long long C(int n, int m) {
  return f[n] * inv[m] % mod * inv[n - m] % mod;
}
int main() {
  cin.tie(0);
  cout.tie(0);
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    long long l, r;
    cin >> l >> r;
    if (r >= l * 2) {
      cout << l << ' ' << 2 * l << endl;
    } else
      cout << -1 << ' ' << -1 << endl;
  }
  return 0;
}

```