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 CPP):

Solution:
/*@author piyush0120*/
/*check whether it has test_cases*/
/*'a'-96 gives 1, similarly '1'-0 gives 1*/
#define pb push_back
#define F first
#define S second
#define lb lower_bound
#define ub upper_bound
#define MP make_pair
#define tos to_string
#define ll long long
#define ull unsigned long long
#define len(a) a.size()
#define PI pair<int,int>
#define MM(a , x) memset(a , x , sizeof(a))
#define ign ignore(numeric_limits<streamsize>::max(), '\n')
#define forl(i,b,n) for(int i=b;i<n;i++)
#define testc int t;cin>>t;while(t--)
#define maxN 1000001
#define MOD 1000000007
#include <bits/stdc++.h>
using namespace std;
using namespace chrono;
int gcd(int a, int b) { if (a == 0) return b; return gcd(b % a, a);}
bool sortByVal(const pair<int, int> &a, const pair<int, int> &b) {return (a.second < b.second);/*sort(v.begin(), v.end(), sortByVal);*/}
/*****************/
void solve()
{
	ll n, a, b; cin >> n >> a >> b;
	if ((n - 1) % b == 0)
	{
		cout << "Yes\n"; return;
	}
	if (a == 1)
	{
		cout << "No\n"; return;
	}
	ll x = 1;
	while (x <= n)
	{
		if ((n - x) % b == 0)
		{
			//cout << x << "," << n - x << "\n";
			cout << "Yes\n"; return;
		}
		x *= a;
	}
	cout << "No\n";
	return;
}

int main()
{

#ifndef ONLINE_JUDGE

	freopen("input2.txt", "r", stdin);
	freopen("output2.txt", "w", stdout);
	freopen("error.txt", "w", stderr);
#endif

	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	cout << fixed;
	cout << setprecision(10);
	auto start = high_resolution_clock::now();
	testc
	solve();
	auto stop = high_resolution_clock::now();
	auto duration = duration_cast<chrono::microseconds>(stop - start);

#ifndef ONLINE_JUDGE
	cerr << "Time: " << duration.count() / 1000 << endl;
#endif
}
/*!-Solve A & B quickly with less submissions-!*/
/*!--Reach -->Pupil--!*/