#include <bits/stdc++.h>
using namespace std;
template <typename T>
T modpow(T base, T exp, T modulus) {
  base %= modulus;
  T result = 1;
  while (exp > 0) {
    if (exp & 1) result = (result * base) % modulus;
    base = (base * base) % modulus;
    exp >>= 1;
  }
  return result;
}
bool sortbysec(const pair<long long, long long> &a,
               const pair<long long, long long> &b) {
  return a.second < b.second;
}
struct cmp {
  bool operator()(const pair<int, int> &a, const pair<int, int> &b) const {
    int lena = a.second - a.first + 1;
    int lenb = b.second - b.first + 1;
    if (lena == lenb) return a.first < b.first;
    return lena > lenb;
  }
};
struct c_h {
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    x ^= FIXED_RANDOM;
    return x ^ (x >> 16);
  }
};
long long calcpow(long long x, long long y) {
  long long temp;
  if (y == 0) return 1;
  temp = calcpow(x, y / 2);
  if (y % 2 == 0)
    return temp * temp;
  else {
    if (y > 0)
      return x * temp * temp;
    else
      return (temp * temp) / x;
  }
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long egcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long x1;
  long long y1;
  long long g = egcd(b, a % b, x1, y1);
  x = y1;
  y = x1 - y1 * (a / b);
  return g;
}
bool isprime(long long a) {
  bool flag = 1;
  for (long long i = 2; i * i <= a; i++) {
    if (a % i == 0) {
      flag = 0;
      break;
    }
  }
  return flag;
}
long long dx[8] = {1, 0, -1, 0, 1, -1, 1, -1};
long long dy[8] = {0, 1, 0, -1, 1, -1, -1, 1};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long x, y;
    cin >> x >> y;
    long long a = -1, b = -1;
    if (y >= 2 * x) {
      a = x;
      b = 2 * x;
    }
    cout << a << " " << b << endl;
  }
  return 0;
}

### Prompt

Please formulate a Cpp solution to the following problem:

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;
template <typename T>
T modpow(T base, T exp, T modulus) {
  base %= modulus;
  T result = 1;
  while (exp > 0) {
    if (exp & 1) result = (result * base) % modulus;
    base = (base * base) % modulus;
    exp >>= 1;
  }
  return result;
}
bool sortbysec(const pair<long long, long long> &a,
               const pair<long long, long long> &b) {
  return a.second < b.second;
}
struct cmp {
  bool operator()(const pair<int, int> &a, const pair<int, int> &b) const {
    int lena = a.second - a.first + 1;
    int lenb = b.second - b.first + 1;
    if (lena == lenb) return a.first < b.first;
    return lena > lenb;
  }
};
struct c_h {
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    x ^= FIXED_RANDOM;
    return x ^ (x >> 16);
  }
};
long long calcpow(long long x, long long y) {
  long long temp;
  if (y == 0) return 1;
  temp = calcpow(x, y / 2);
  if (y % 2 == 0)
    return temp * temp;
  else {
    if (y > 0)
      return x * temp * temp;
    else
      return (temp * temp) / x;
  }
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long egcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long x1;
  long long y1;
  long long g = egcd(b, a % b, x1, y1);
  x = y1;
  y = x1 - y1 * (a / b);
  return g;
}
bool isprime(long long a) {
  bool flag = 1;
  for (long long i = 2; i * i <= a; i++) {
    if (a % i == 0) {
      flag = 0;
      break;
    }
  }
  return flag;
}
long long dx[8] = {1, 0, -1, 0, 1, -1, 1, -1};
long long dy[8] = {0, 1, 0, -1, 1, -1, -1, 1};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long x, y;
    cin >> x >> y;
    long long a = -1, b = -1;
    if (y >= 2 * x) {
      a = x;
      b = 2 * x;
    }
    cout << a << " " << b << endl;
  }
  return 0;
}

```