Given two integers **L** and **R**, the task is to find a pair of integers from the range **[L, R]** having LCM within the range [L, R] as well. If no such pair can be obtained, then print **-1**. If multiple pairs exist, print any one of them.

**Examples:**

Input:L =13, R = 69Output:X =13, Y = 26Explanation:LCM(x, y) = 26 which satisfies the conditions L ≤ x < y ≤ R and L <= LCM(x, y) <= R

Input:L =1, R = 665Output:X = 1, Y = 2

**Naive Approach: **The simplest approach is to generate every pair between **L** and **R** and compute their LCM. Print a pair having LCM between the range **L** and **R**. If no pair is found to have LCM in the given range, print **“-1”**.

**Time Complexity:** O(N)**Auxiliary Space:** O(1)

**Efficient Approach: **The problem can be solved by using** **the Greedy technique based on the observation that **LCM(x, y)** is at least equal to **2*x** which is LCM of **(x, 2*x)**. Below are the steps to implement the approach:

- Select the value of x as L and compute the value of y as 2*x
- Check if y is less than R or not.
- If y is less than R then print the pair (x, y)
- Else print “-1”

Below is the implementation of the above approach:

`// C++ implementation of the above approach` `#include <iostream>` `using` `namespace` `std;`
`void` `lcmpair(` `int` `l, ` `int` `r)`
`{` ` ` `int` `x, y;`
` ` `x = l;`
` ` `y = 2 * l;`
` ` `// Checking if any pair is possible`
` ` `// or not in range(l, r)`
` ` `if` `(y > r) {`
` ` `// If not possible print(-1)`
` ` `cout << ` `"-1\n"` `;`
` ` `}`
` ` `else` `{`
` ` `// Print LCM pair`
` ` `cout << ` `"X = "` `<< x << ` `" Y = "`
` ` `<< y << ` `"\n"` `;`
` ` `}`
`}` `// Driver code` `int` `main()`
`{` ` ` `int` `l = 13, r = 69;`
` ` `// Function call`
` ` `lcmpair(l, r);`
` ` `return` `0;`
`}` |

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`// Java implementation of the above approach` `import` `java.util.*;`
`class` `GFG{`
`static` `void` `lcmpair(` `int` `l, ` `int` `r)`
`{` ` ` `int` `x, y;`
` ` `x = l;`
` ` `y = ` `2` `* l;`
` ` `// Checking if any pair is possible`
` ` `// or not in range(l, r)`
` ` `if` `(y > r) `
` ` `{`
` ` ` ` `// If not possible print(-1)`
` ` `System.out.print(` `"-1\n"` `);`
` ` `}`
` ` `else`
` ` `{`
` ` ` ` `// Print LCM pair`
` ` `System.out.print(` `"X = "` `+ x + `
` ` `" Y = "` `+ y + ` `"\n"` `);`
` ` `}`
`}` `// Driver code` `public` `static` `void` `main(String[] args)`
`{` ` ` `int` `l = ` `13` `, r = ` `69` `;`
` ` `// Function call`
` ` `lcmpair(l, r);`
`}` `}` `// This code is contributed by 29AjayKumar ` |

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`# Python3 implementation of the above approach ` `def` `lcmpair(l, r):`
` ` `x ` `=` `l`
` ` `y ` `=` `2` `*` `l`
` ` `# Checking if any pair is possible`
` ` `# or not in range(l, r)`
` ` `if` `(y > r):`
` ` `# If not possible print(-1)`
` ` `print` `(` `-` `1` `)`
` ` `else` `:`
` ` ` ` `# Print LCM pair`
` ` `print` `(` `"X = {} Y = {}"` `.` `format` `(x, y))`
`# Driver Code` `l ` `=` `13`
`r ` `=` `69`
`# Function call` `lcmpair(l, r)` `# This code is contributed by Shivam Singh` |

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`// C# implementation of the above approach` `using` `System;`
`class` `GFG{`
`static` `void` `lcmpair(` `int` `l, ` `int` `r)`
`{` ` ` `int` `x, y;`
` ` `x = l;`
` ` `y = 2 * l;`
` ` `// Checking if any pair is possible`
` ` `// or not in range(l, r)`
` ` `if` `(y > r) `
` ` `{ `
` ` `// If not possible print(-1)`
` ` `Console.Write(` `"-1\n"` `);`
` ` `}`
` ` `else`
` ` `{ `
` ` `// Print LCM pair`
` ` `Console.Write(` `"X = "` `+ x + `
` ` `" Y = "` `+ y + ` `"\n"` `);`
` ` `}`
`}` `// Driver code` `public` `static` `void` `Main(String[] args)`
`{` ` ` `int` `l = 13, r = 69;`
` ` `// Function call`
` ` `lcmpair(l, r);`
`}` `}` `// This code is contributed by shikhasingrajput` |

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**Output:**

X = 13 Y = 26

**Time Complexity:** O(1)**Auxiliary Space:** O(1)

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