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 PYTHON3):

Solution:
#import math,statistics,random

# STEPS FOR B- Plus and Multiply
# write any random term of infinite set and simplify it you will get a
# general term i.e. a**(max power of a possible) + b*(function of a).
# so to achieve n if we can go from n to 1 backwards then n is in the set.
# so we can subtract b till it gets 0 and then divide by a till it becomes 1.But
# this will give us TLE.
# so we will first subtract by a then check if its divisible by b. if not then 
# subtract by a**2/a**3/a**4,...,etc till it does not get divisible by b.
# as we are subtracting by exponetial elements it will take much less time.
# but for a==1 we do it seperately as in that case remainder=1(always)
# but the problem when a==1 and n%b==1 is that if a=b=n=1 then n%b==0
# so it will print NO but if we see n=1 is always possible.Thus (n-1)%b==0
# is a valid option.

def void():
    n,a,b=map(int,input().split())
    if a==1:
        if (n-1)%b==0:
            print("Yes")
        else:
            print("No")
    else:
        i=1
        valid=False
        while i<=n:
            if (n-i)%b==0:
                valid=True
                break
            i*=a
        if valid:
            print("Yes")
        else:
            print("No")
        
for i in range(int(input())):
    void()