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:
// Muhammad Mirza Fathan 2021
#include <bits/stdc++.h>
using namespace std;

/************************/
/* TEMPLATES */

#define fr front
#define bk back
#define pb push_back
#define mp make_pair
#define sc second
#define fi first
# define ll long long
# define ld long double
# define rep(i,x,y) for(ll i=x;i<y;i++)
# define rrep(i,x,y) for(ll i=x;i>=y;i--)
# define debug_stl(x) for(auto i:x){cout<<i<<" ";}cout<<endl;
# define debug_1d(x,a,b) rep(i,a,b){cout<<x[i]<<" ";}cout<<endl;
# define debug_2d(x,a1,b1,a2,b2) rep(i,a1,b1){rep(j,a2,b2){cout<<x[i][j]<<" ";}cout<<endl;}
# define all(x) x.begin(), x.end()
# define MOD (ll)(1e9+7)
# define INF8 (ll)(1e17+5)
# define endl '\n'

typedef vector<ll> VI;
typedef vector<bool> VB;
typedef vector<string> VS;
typedef vector<vector<ll> > VVI;
typedef pair<ll,ll> PI;
typedef pair<ll,PI > PPI;
typedef vector<PI> VP;
typedef vector<PPI> VPP;

vector<ll> primeList;
vector<bool> primes;


ll mod(ll x) {
  return (x%MOD + MOD)%MOD;
}

ll power(ll x, ll y) {
  ll res = 1;
  x = x%MOD;
  if (x == 0) return 0;
  while (y > 0) {
    if (y&1) res = mod(res * x);
    y >>= 1; x = mod(x * x);
  }
  return res;
}

ll gcd(ll a, ll b)
{
    if (a == 0)
      return b;
    if (b == 0)
      return a;

    if (a == b)
      return a;
    if (a > b)
        return gcd(a%b, b);
    return gcd(a, b%a);
}

void sieve() {
  primes[0] = false;
  primes[1] = false;
  for(ll i=2; i<primes.size(); i++) primes[i] = true;

  for(ll i=2; i*i<primes.size(); i++) {
    if(primes[i]) {
      for(ll j=i*i; j<primes.size(); j+=i)
        primes[j] = false;
    }
  }
  for(ll i=2; i<primes.size(); i++) {
    if(primes[i]) primeList.push_back(i);
  }

}

bool isPrime(ll n)
{
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;

    if (n % 2 == 0 || n % 3 == 0)
        return false;

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

    return true;
}


/* From hu_tao:

        Random stuff to try when stuck:
           -if it's 2C then it's dp
           -for combo/probability problems, expand the given form we're interested in
           -make everything the same then build an answer (constructive, make everything 0 then do something)
           -something appears in parts of 2 --> model as graph
           -assume a greedy then try to show why it works
           -find way to simplify into one variable if multiple exist
           -treat it like fmc (note any passing thoughts/algo that could be used so you can revisit them)
           -find lower and upper bounds on answer
           -figure out what ur trying to find and isolate it
           -see what observations you have and come up with more continuations
           -work backwards (in constructive, go from the goal to the start)
           -turn into prefix/suffix sum argument (often works if problem revolves around adjacent array elements)
           -instead of solving for answer, try solving for complement (ex, find n-(min) instead of max)
           -draw something
           -simulate a process
           -dont implement something unless if ur fairly confident its correct
           -after 3 bad submissions move on to next problem if applicable
           -do something instead of nothing and stay organized
           -write stuff down
        Random stuff to check when wa:
           -if code is way too long/cancer then reassess
           -switched N/M
           -int overflow
           -switched variables
           -wrong MOD
           -hardcoded edge case incorrectly
        Random stuff to check when tle:
           -continue instead of break
           -condition in for/while loop bad
        Random stuff to check when rte:
           -switched N/M
           -long to int/int overflow
           -division by 0
           -edge case for empty list/data structure/N=1
     */

/************************/

pair<ll,ll> logarithm(ll a, ll b) {
  ll ans = 0;
  ll temp =a;
  while(a<=b) {
    a*=temp;
    ans++;
  }
  return mp(ans,a/temp);
}

void solve() {
  ll n, a, b; cin >> n >> a >> b;

  if(b==1) {
    cout << "Yes" << endl;
    return;
  }
  if(a==1) {
    if((n-a)%b==0) cout << "YES" << endl;
    else cout << "NO" << endl;
    return;
  }

  pair<ll,ll> an = logarithm(a,n);
  ll x = an.first;
  ll ax = an.second;

  while(ax>0) {
    ll c = n-ax;
    if(c%b==0) {
      cout << "Yes" << endl;
      return;
    }
    ax/=a;
//    cout << ax << endl;
  }

  cout << "No" << endl;


}

int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);

  ll t; cin >> t;
  while(t--) {
    solve();
  }

  return 0;
}