Given two integers **N** and **K**, the task is to represent **N** as sum of **K** odd numbers. If it is not possible to create the sum then output **-1**.

**Note:** The representation may contain duplicate odd numbers.

**Examples:**

Input:N = 5, K = 3

Output:1, 1, 3

Explanation:

The given number N can be represented as 1 + 1 + 3 = 5

Input:N = 7, K = 5

Output:1, 1, 1, 1, 3

Explanation:

The given number N can be represented as 1 + 1 + 1 + 1 + 3 = 7

**Approach:**

To solve the problem mentioned above a simple solution is to **maximise the occurrence of 1** which is the possible smallest odd number. Necessary conditions for representing the number N as K odd numbers are:

**(K – 1)**must be less than N.**N – (K – 1)**must be a Odd number.

Below is the implementation of the above approach:

## C++

`// C++ implementation to represent ` `// N as sum of K even numbers ` ` ` `#include <bits/stdc++.h> ` ` ` `using` `namespace` `std; ` ` ` `// Function to print the representation ` `void` `sumOddNumbers(` `int` `N, ` `int` `K) ` `{ ` ` ` `int` `check = N - (K - 1); ` ` ` ` ` `// N must be greater than equal to 2*K ` ` ` `// and must be odd ` ` ` `if` `(check > 0 && check % 2 == 1) { ` ` ` `for` `(` `int` `i = 0; i < K - 1; i++) { ` ` ` `cout << ` `"1 "` `; ` ` ` `} ` ` ` `cout << check; ` ` ` `} ` ` ` `else` ` ` `cout << ` `"-1"` `; ` `} ` ` ` `// Driver Code ` `int` `main() ` `{ ` ` ` ` ` `int` `N = 5; ` ` ` `int` `K = 3; ` ` ` ` ` `sumOddNumbers(N, K); ` ` ` `return` `0; ` `} ` |

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## Java

`// Java implementation to represent ` `// N as sum of K even numbers ` `import` `java.util.*; ` ` ` `class` `GFG{ ` ` ` `// Function to print the representation ` `static` `void` `sumOddNumbers(` `int` `N, ` `int` `K) ` `{ ` ` ` `int` `check = N - (K - ` `1` `); ` ` ` ` ` `// N must be greater than equal ` ` ` `// to 2*K and must be odd ` ` ` `if` `(check > ` `0` `&& check % ` `2` `== ` `1` `) ` ` ` `{ ` ` ` `for` `(` `int` `i = ` `0` `; i < K - ` `1` `; i++) ` ` ` `{ ` ` ` `System.out.print(` `"1 "` `); ` ` ` `} ` ` ` `System.out.print(+check); ` ` ` `} ` ` ` `else` ` ` `System.out.println(` `"-1 "` `); ` `} ` ` ` `// Driver Code ` `public` `static` `void` `main(String args[]) ` `{ ` ` ` `int` `N = ` `5` `; ` ` ` `int` `K = ` `3` `; ` ` ` ` ` `sumOddNumbers(N, K); ` `} ` `} ` ` ` `// This code is contributed by AbhiThakur ` |

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## Python3

`# Python3 implementation to represent ` `# N as sum of K even numbers ` ` ` `# Function to print the representation ` `def` `sumOddNumbers(N, K): ` ` ` ` ` `check ` `=` `N ` `-` `(K ` `-` `1` `) ` ` ` ` ` `# N must be greater than equal ` ` ` `# to 2*K and must be odd ` ` ` `if` `(check > ` `0` `and` `check ` `%` `2` `=` `=` `1` `): ` ` ` `for` `i ` `in` `range` `(` `0` `, K ` `-` `1` `): ` ` ` `print` `(` `"1"` `, end ` `=` `" "` `) ` ` ` ` ` `print` `(check, end ` `=` `" "` `) ` ` ` ` ` `else` `: ` ` ` `print` `(` `"-1"` `) ` ` ` `# Driver Code ` `N ` `=` `5` `K ` `=` `3` `; ` ` ` `sumOddNumbers(N, K) ` ` ` `# This code is contributed by PratikBasu ` |

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## C#

`// C# implementation to represent ` `// N as sum of K even numbers ` `using` `System; ` ` ` `class` `GFG{ ` ` ` `// Function to print the representation ` `static` `void` `sumOddNumbers(` `int` `N, ` `int` `K) ` `{ ` ` ` `int` `check = N - (K - 1); ` ` ` ` ` `// N must be greater than equal ` ` ` `// to 2*K and must be odd ` ` ` `if` `(check > 0 && check % 2 == 1) ` ` ` `{ ` ` ` `for` `(` `int` `i = 0; i < K - 1; i++) ` ` ` `{ ` ` ` `Console.Write(` `"1 "` `); ` ` ` `} ` ` ` `Console.Write(+check); ` ` ` `} ` ` ` `else` ` ` `Console.WriteLine(` `"-1 "` `); ` `} ` ` ` `// Driver Code ` `public` `static` `void` `Main() ` `{ ` ` ` `int` `N = 5; ` ` ` `int` `K = 3; ` ` ` ` ` `sumOddNumbers(N, K); ` `} ` `} ` ` ` `// This code is contributed by Code_Mech ` |

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**Output:**

1 1 3

**Time Complexity:** O(K)

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