Problem: There is an infinite set generated as follows:

  * 1 is in this set. 
  * If x is in this set, x ⋅ a and x+b both are in this set. 



For example, when a=3 and b=6, the five smallest elements of the set are:

  * 1, 
  * 3 (1 is in this set, so 1⋅ a=3 is in this set), 
  * 7 (1 is in this set, so 1+b=7 is in this set), 
  * 9 (3 is in this set, so 3⋅ a=9 is in this set), 
  * 13 (7 is in this set, so 7+b=13 is in this set). 



Given positive integers a, b, n, determine if n is in this set.

Input

The input consists of multiple test cases. The first line contains an integer t (1≤ t≤ 10^5) — the number of test cases. The description of the test cases follows.

The only line describing each test case contains three integers n, a, b (1≤ n,a,b≤ 10^9) separated by a single space.

Output

For each test case, print "Yes" if n is in this set, and "No" otherwise. You can print each letter in any case.

Example

Input


5
24 3 5
10 3 6
2345 1 4
19260817 394 485
19260817 233 264


Output


Yes
No
Yes
No
Yes

Note

In the first test case, 24 is generated as follows:

  * 1 is in this set, so 3 and 6 are in this set; 
  * 3 is in this set, so 9 and 8 are in this set; 
  * 8 is in this set, so 24 and 13 are in this set. 



Thus we can see 24 is in this set.

The five smallest elements of the set in the second test case is described in statements. We can see that 10 isn't among them.

Solution (in JAVA):

Solution:

import java.io.*;
import java.util.*;
public class Main
{
	public static void main(String[] args)throws Exception
	{
	    Main ob=new Main();
	    ob.fun();
	}
	public void fun()throws Exception
	{
	    BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
	    PrintWriter pw=new PrintWriter(System.out);
	    int t=Integer.parseInt(br.readLine());
	    while(t-->0)
	    {
	        String s[]=(br.readLine()).split(" ");
	        long n=Long.parseLong(s[0]);
	        long a=Long.parseLong(s[1]);
	        long b=Long.parseLong(s[2]);
	        long aPow=1l;
	        if(a==1)
	        {
	            if((n-1l)%b==0)
	            {
	                pw.println("Yes");
	            }
	            else
	            {
	                pw.println("No");
	            }
	            continue;
	        }
	        boolean ok=false;
	        while(aPow<=n)
	        {
	            if(((n-aPow)%b)==0)
	            {
	                ok=true;
	                break;
	            }
	            aPow*=a;
	        }
	        if(ok)
	        {
	            pw.println("Yes");
	        }
	        else
	        {
	            pw.println("No");
	        }
	    }
	    pw.flush();
	}
	public int floorSearch(int arr[], int low,int high, int x)
    {
        // If low and high cross each other
        if (low > high)
            return -1;
 
        // If last element is smaller than x
        if (x >= arr[high])
            return high;
 
        // Find the middle point
        int mid = (low + high) / 2;
 
        // If middle point is floor.
        if (arr[mid] == x)
            return mid;
 
        // If x lies between mid-1 and mid
        if (
            mid > 0 && arr[mid - 1] <= x && x < arr[mid])
            return mid - 1;
 
        // If x is smaller than mid, floor
        // must be in left half.
        if (x < arr[mid])
            return floorSearch(
                arr, low,
                mid - 1, x);
 
        // If mid-1 is not floor and x is
        // greater than arr[mid],
        return floorSearch(
            arr, mid + 1, high,
            x);
    }
	public int ceilSearch(int arr[], int low, int high, int x)
    {
      int mid;   
       
      /* If x is smaller than or equal to the
         first element, then return the first element */
      if(x <= arr[low])
        return low;
      
      /* If x is greater than the last
         element, then return -1 */
      if(x > arr[high])
        return -1; 
      
      /* get the index of middle element
         of arr[low..high]*/
      mid = (low + high)/2;  /* low + (high - low)/2 */
      
      /* If x is same as middle element,
         then return mid */
      if(arr[mid] == x)
        return mid;
          
      /* If x is greater than arr[mid], then
         either arr[mid + 1] is ceiling of x or
         ceiling lies in arr[mid+1...high] */
      else if(arr[mid] < x)
      {
        if(mid + 1 <= high && x <= arr[mid+1])
          return mid + 1;
        else
          return ceilSearch(arr, mid+1, high, x);
      }
      
      /* If x is smaller than arr[mid],
         then either arr[mid] is ceiling of x
         or ceiling lies in arr[low...mid-1] */  
      else
      {
        if(mid - 1 >= low && x > arr[mid-1])
          return mid;
        else   
          return ceilSearch(arr, low, mid - 1, x);
      }
    }
	
}