Problem: Programmer Rostislav got seriously interested in the Link/Cut Tree data structure, which is based on Splay trees. Specifically, he is now studying the expose procedure.

Unfortunately, Rostislav is unable to understand the definition of this procedure, so he decided to ask programmer Serezha to help him. Serezha agreed to help if Rostislav solves a simple task (and if he doesn't, then why would he need Splay trees anyway?)

Given integers l, r and k, you need to print all powers of number k within range from l to r inclusive. However, Rostislav doesn't want to spent time doing this, as he got interested in playing a network game called Agar with Gleb. Help him!

Input

The first line of the input contains three space-separated integers l, r and k (1 ≤ l ≤ r ≤ 1018, 2 ≤ k ≤ 109).

Output

Print all powers of number k, that lie within range from l to r in the increasing order. If there are no such numbers, print "-1" (without the quotes).

Examples

Input

1 10 2


Output

1 2 4 8 

Input

2 4 5


Output

-1

Note

Note to the first sample: numbers 20 = 1, 21 = 2, 22 = 4, 23 = 8 lie within the specified range. The number 24 = 16 is greater then 10, thus it shouldn't be printed.

Solution (in JAVA):

Solution:
import java.io.*;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.util.*;
import java.util.Comparator;

public class scratch_25 {
    static class Reader {
        static BufferedReader reader;
        static StringTokenizer tokenizer;

        /** call this method to initialize reader for InputStream */
        static void init(InputStream input) {
            reader = new BufferedReader(
                    new InputStreamReader(input) );
            tokenizer = new StringTokenizer("");
        }

        /** get next word */
        static String next() throws IOException {
            while ( ! tokenizer.hasMoreTokens() ) {
                //TODO add check for eof if necessary
                tokenizer = new StringTokenizer(
                        reader.readLine() );
            }
            return tokenizer.nextToken();
        }

        static int nextInt() throws IOException {
            return Integer.parseInt( next() );
        }

        static double nextDouble() throws IOException {
            return Double.parseDouble( next() );
        }
        static long nextLong() throws IOException{
            return Long.parseLong(next());
        }
    }






    static class ans{
        int a;
        int b;
        public ans(int a, int b){
            this.a=a;
            this.b=b;

        }
    }
    static class sort implements Comparator<ans>{


        @Override
        public int compare(ans a1,ans a2){
            return a1.a-a2.a;
        }




    }
   









    public static void main(String[] args) throws IOException {
        Reader.init(System.in);
        long start= Reader.nextLong();
        long end= Reader.nextLong();
        long num= Reader.nextLong();
        BigInteger n= BigInteger.valueOf(num);
        BigInteger st= BigInteger.valueOf(start);
        BigInteger ed= BigInteger.valueOf(end);
        BigInteger f= BigInteger.valueOf(1);

        while(f.compareTo(st)<0){
            f=f.multiply(n);
        }
        if(f.compareTo(ed)>0){
            System.out.println(-1);
            System.exit(0);
        }
        while(f.compareTo(ed)<=0){
            System.out.print(f+" ");
            f=f.multiply(n);
        }

































    }

    public static char[] getCharArray(char[] array) {
        String _array = "";
        for(int i = 0; i < array.length; i++) {
            if(_array.indexOf(array[i]) == -1) // check if a char already exist, if not exist then return -1
                _array = _array+array[i];      // add new char
        }
        return _array.toCharArray();
    }










    static void merge(long arr[], int l, int m, int r)
    {
        // Find sizes of two subarrays to be merged
        int n1 = m - l + 1;
        int n2 = r - m;

        /* Create temp arrays */
        long L[] = new long [n1];
        long R[] = new long [n2];

        /*Copy data to temp arrays*/
        for (int i=0; i<n1; ++i)
            L[i] = arr[l + i];
        for (int j=0; j<n2; ++j)
            R[j] = arr[m + 1+ j];


        /* Merge the temp arrays */

        // Initial indexes of first and second subarrays
        int i = 0, j = 0;

        // Initial index of merged subarry array
        int k = l;
        while (i < n1 && j < n2)
        {
            if (L[i] <= R[j])
            {
                arr[k] = L[i];
                i++;
            }
            else
            {
                arr[k] = R[j];
                j++;
            }
            k++;
        }

        /* Copy remaining elements of L[] if any */
        while (i < n1)
        {
            arr[k] = L[i];
            i++;
            k++;
        }

        /* Copy remaining elements of R[] if any */
        while (j < n2)
        {
            arr[k] = R[j];
            j++;
            k++;
        }
    }

    // Main function that sorts arr[l..r] using
    // merge()
    static void sort(long arr[], int l, int r)
    {
        if (l < r)
        {
            // Find the middle point
            int m = (l+r)/2;

            // Sort first and second halves
            sort(arr, l, m);
            sort(arr , m+1, r);

            // Merge the sorted halves
            merge(arr, l, m, r);
        }
    }
    static boolean ans(int check, long arr[],int n,int med,int k){
        long ans=0;
        for (int i = med; i <n ; i++) {
            if(check>=arr[i]){
                ans+=check-arr[i];
            }
        }
        return ans<=k;




    }


    static boolean isPrime(long n)
    {
        // Corner cases
        if (n <= 1) return false;
        if (n <= 3) return true;

        // This is checked so that we can skip
        // middle five numbers in below loop
        if (n % 2 == 0 || n % 3 == 0) return false;

        for (long i = 5; i * i <= n; i = i + 6)
            if (n % i == 0 || n % (i + 2) == 0)
                return false;

        return true;
    }






    static   int partition(double arr[],int arr1[],int arr2[], int low, int high)
    {
        double pivot = arr[high];
        int i = (low-1); // index of smaller element
        for (int j=low; j<high; j++)
        {
            // If current element is smaller than the pivot
            if (arr[j] < pivot)
            {
                i++;

                // swap arr[i] and arr[j]
                double temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
                int temp1=arr1[i];
                arr1[i]=arr1[j];
                arr1[j]=temp1;
                int temp2=arr2[i];
                arr2[i]=arr2[j];
                arr2[j]=temp2;

            }
        }

        // swap arr[i+1] and arr[high] (or pivot)
        double temp = arr[i+1];
        arr[i+1] = arr[high];
        arr[high] = temp;
        int temp1= arr1[i+1];
        arr1[i+1]=arr1[high];
        arr1[high]= temp1;
        int temp2= arr2[i+1];
        arr2[i+1]=arr2[high];
        arr2[high]= temp2;

        return i+1;
    }


    /* The main function that implements QuickSort()
      arr[] --> Array to be sorted,
      low  --> Starting index,
      high  --> Ending index */
    static void sort(double arr[],int arr1[],int arr2[], int low, int high)
    {
        if (low < high)
        {
            /* pi is partitioning index, arr[pi] is
              now at right place */
            int pi = partition(arr,arr1,arr2, low, high);

            // Recursively sort elements before
            // partition and after partition
            sort(arr,arr1,arr2, low, pi-1);
            sort(arr,arr1, arr2,pi+1, high);
        }
    }
    static int removeDuplicates(int arr[], int n)
    {
        // Return, if array is empty
        // or contains a single element
        if (n==0 || n==1)
            return n;

        int[] temp = new int[n];

        // Start traversing elements
        int j = 0;
        for (int i=0; i<n-1; i++)
            // If current element is not equal
            // to next element then store that
            // current element
            if (arr[i] != arr[i+1])
                temp[j++] = arr[i];

        // Store the last element as whether
        // it is unique or repeated, it hasn't
        // stored previously
        temp[j++] = arr[n-1];

        // Modify original array
        for (int i=0; i<j; i++)
            arr[i] = temp[i];

        return j;
    }
    static long maxPrimeFactors( long n)
    {
        // Initialize the maximum prime
        // factor variable with the
        // lowest one
        long maxPrime = -1;

        // Print the number of 2s
        // that divide n
        while (n % 2 == 0) {
            maxPrime = 2;

            // equivalent to n /= 2
            n >>= 1;
        }

        // n must be odd at this point,
        // thus skip the even numbers
        // and iterate only for odd
        // integers
        for (int i = 3; i <= Math.sqrt(n); i += 2) {
            while (n % i == 0) {
                maxPrime = i;
                n = n / i;
            }
        }

        // This condition is to handle
        // the case when n is a prime
        // number greater than 2
        if (n > 2)
            maxPrime = n;

        return maxPrime;
    }




}