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;
typedef long long ll;
#define mod 1000000007
#define inf 1e17
#define mod 1000000007
#define INF 1e17
#define eps 1e-9
#define FASTIO ios_base::sync_with_stdio(false);cin.tie(NULL);
#define MAXN   1000001
 
// stores smallest prime factor for every number
ll spf[MAXN];
 
// Calculating SPF (Smallest Prime Factor) for every
// number till MAXN.
// Time Complexity : O(nloglogn)
void sieve()
{
    spf[1] = 1;
    for (ll i=2; i<MAXN; i++)
 
        // marking smallest prime factor for every
        // number to be itself.
        spf[i] = i;
 
    // separately marking spf for every even
    // number as 2
    for (ll i=4; i<MAXN; i+=2)
        spf[i] = 2;
 
    for (ll i=3; i*i<MAXN; i++)
    {
        // checking if i is prime
        if (spf[i] == i)
        {
            // marking SPF for all numbers divisible by i
            for (ll j=i*i; j<MAXN; j+=i)
 
                // marking spf[j] if it is not
                // previously marked
                if (spf[j]==j)
                    spf[j] = i;
        }
    }
}
 
// A O(log n) function returning primefactorization
// by dividing by smallest prime factor at every step
vector<ll > getFactorization(ll x)
{
    vector<ll> ret;
    while (x != 1)
    {
        ret.push_back(spf[x]);
        x = x / spf[x];
    }
    return ret;
}
ll power(ll x,ll y)
{
    ll res = 1;     // Initialize result
 
    x = x % mod; // Update x if it is more than or
                // equal to p
  
    if (x == 0) return 0; // In case x is divisible by p;
 
    while (y > 0)
    {
        // If y is odd, multiply x with result
        if (y & 1)
            res = (res*x) % (mod);
 
        // y must be even now
        y = y>>1; // y = y/2
        x = (x*x) % (mod);
    }
    return res;
}

class myComparator
{
public:
    bool operator() (ll p1, ll p2)
    {
        return p1 > p2;
    }
};

map<string,ll>mp2;
map<ll,string>mp3;
bool compa(vector<ll>a,vector<ll>b)
{
if (a[0]==b[0]){
  if (a[1]==b[1]){
    if (a[2]==b[2]){
    return mp3[a[3]]<mp3[b[3]];
    }
    return a[2]>b[2];
  }
  return a[1]>b[1];
}
return a[0]>b[0];
}


void fun(){
ll i,j,k,t,r,h,m,n,a,b;
cin>>n>>a>>b;
ll x=n;
ll d=1;

for(i=0;i<33;i++){
  if (d>n){
    break;
  }
  if ((n-d)%b==0){
    cout<<"Yes"<<endl;
    return;
  }
  d=a*d;
  if (d==1)
  break;
}

cout<<"No"<<endl;return;

}


int main(){
  ll t=1;
  FASTIO;
  cin>>t;
  while(t--){
    fun();
  }
}