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:
#include <bits/stdc++.h>
using namespace std;

#define ms(s, n) memset(s, n, sizeof(s))
#define f(i, a, b) for (int i = (a); i <=(b); i++)
#define FORd(i, a, b) for (int i = (a) - 1; i >= (b); i--)
#define FORall(it, a) for (__typeof((a).begin()) it = (a).begin(); it != (a).end(); it++)
#define sz(a) int((a).size())
#define present(t, x) (t.find(x) != t.end())
#define all(a) (a).begin(), (a).end()
#define uni(a) (a).erase(unique(all(a)), (a).end())
#define pb push_back
#define pf push_front
#define mp make_pair
#define fi first
#define se second
#define prec(n) fixed<<setprecision(n)
#define bit(n, i) (((n) >> (i)) & 1)
#define bitcount(n) __builtin_popcountll(n)
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int, int> pi;
typedef vector<ll> vi;
typedef vector<pi> vii;
const int MOD = (int) 1e9 + 7;
const int MOD2 = MOD + 2; //1007681537;
const int INF = (int) 1e9;
const ll LINF = (ll) 1e18;
const ld PI = acos((ld) - 1);
const ld EPS = 1e-9;
inline ll gcd(ll a, ll b) {ll r; while (b) {r = a % b; a = b; b = r;} return a;}
inline ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
inline ll fpow(ll n, ll k, int p = MOD) {ll r = 1; for (; k; k >>= 1) {if (k & 1) r = r * n % p; n = n * n % p;} return r;}
template<class T> inline int chkmin(T& a, const T& val) {return val < a ? a = val, 1 : 0;}
template<class T> inline int chkmax(T& a, const T& val) {return a < val ? a = val, 1 : 0;}
inline ll isqrt(ll k) {ll r = sqrt(k) + 1; while (r * r > k) r--; return r;}
inline ll icbrt(ll k) {ll r = cbrt(k) + 1; while (r * r * r > k) r--; return r;}
inline void addmod(int& a, int val, int p = MOD) {if ((a = (a + val)) >= p) a -= p;}
inline void submod(int& a, int val, int p = MOD) {if ((a = (a - val)) < 0) a += p;}
inline int mult(int a, int b, int p = MOD) {return (ll) a * b % p;}
inline int inv(int a, int p = MOD) {return fpow(a, p - 2, p);}
inline int sign(ld x) {return x < -EPS ? -1 : x > +EPS;}
inline int sign(ld x, ld y) {return sign(x - y);}
int modulo(int a, int b) { return (a % b + b) % b; }
#define db(x) cerr << #x << " = " << (x) << " ";
#define endln cerr << "\n";

ll loga(ll a, ll b)
{
	return log(a) / log(b);
}
int main()
{
	IOS;
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif
	int t;
	cin >> t;
	while (t--) {
		ll n, a, b;
		cin >> n >> a >> b;
		if (a == 1) {
			if (n == 1) {
				cout << "Yes\n";
			}
			else {
				ll rem = n - 1;
				if (rem % b == 0) {
					cout << "Yes\n";
				}
				else {
					cout << "No\n";
				}
			}
		}
		else {
			ll x = loga(n, a);
			//cout << x << "\n";
			if ((ll)pow(a, x) == n || (ll)pow(a, x + 1) == n) {
				cout << "Yes\n";
			}
			else {
				bool ok = 0;
				for (int i = x + 2; i >= 0; i--) {
					ll rem = n - (ll)pow(a, i);
					if (rem >= 0 && rem % b == 0) {
						cout << "Yes\n"; ok = 1;
						break;
					}
				}
				if (!ok) {
					cout << "No\n";
				}
			}
		}
	}
	// read the stuff at the bottom
	return 0;
}
/* stuff you should look for
        * First , try to think about most basic or brute force solution and then optimise it step by step
	* int overflow, array bounds
	* special cases (n=1?)
	* do smth instead of nothing and stay organized
	* WRITE STUFF DOWN
	* use int when constraints limits are tight
	* Time taken : global variable < pass by refernce < pass by value
	* It takes significant amt. of time in pass by value
	* Avoid Using endl
*/