vector<int> Permutation(vector<int> arr, int k){
    vector<int>result;
    int size = arr.size();
    int start = 0;
    int end = k;
    int count = 0;
    while(start < size){
        swap(arr[start], arr[end]);
        count++;
        if (k != 1){
            start = start + 1;
            end = end + 1;
            if (count == k){
                start = start + k;
                end = end + k;
                count = 0;
            }
        } else {
            if (count == 1){
                start = start + 2*k;
                end = end + 2*k;
                count = 0;
            }
        }
    }
    result = arr;
    return result;
}
vector<int> absolutePermutation(int n, int k) {
    vector<int> newarr;
    vector<int> result;
    int m = n % k;
    if (k == 0){
        for(int i = 0; i < n; i++){newarr.push_back(i+1);}
        result = newarr;
    }
    else if (k == 1)
    {
        if (n % 2 != 0){newarr.push_back(-1);result = newarr;}
        else {
            for (int i = 0; i < n; i++) newarr.push_back(i+1);
            result = Permutation(newarr, k);
        }
    }
    else {
        if (m != 0){newarr.push_back(-1);result = newarr;}
        else 
        {
            if ((n/k) % 2 != 0){newarr.push_back(-1);result = newarr;}
            else {
                for (int i = 0; i < n; i++) newarr.push_back(i+1);
                result = Permutation(newarr, k);}
        }
    }
    return result;
}

def absolutePermutation(n, k):
    # Write your code here    
    if k == 0:
        return [i for i in range(1,n+1,1)]
    elif n % (2*k) != 0:
        return [-1]
    else:
        if k == 1:
            return [i+1 if i%2 != 0 else i-1 for i in range(1,n+1,1)]
        elif k <= n/2 and n % k == 0:
            a = 1
            flag = True
            ans = []
            while a <= n:
                if flag:
                    ans.append(a+k)
                else:
                    ans.append(a-k)
                a += 1
                if a % k == 1:
                    flag = not flag
            return ans
        else:
            return [-1]

function absolutePermutation(n, k) {
  let a = [];
  for (let i = 1; i <= n; i++) {
    a.push(i);
  }
  k = Math.abs(k);
  if (k == 0) return a;
  else if (k > n / 2 + 1 || n % k != 0 || (n/k % 2 != 0)) return [-1];
  else {
    a = [];
    for (let i = 1; i <= n; i++) {
      a.push(i + Math.pow(-1, Math.floor((i - 0.1) / k)) * k);
    }
    return a;
  }
}