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 int long long
#define f(i, b) for (int(i) = 0; (i) < (b); (i)++)
#define fa(i, a, b) for (int(i) = (a); (i) < (b); (i)++)
#define fd(i, a, b) for (int(i) = (a); (i) >= (b); (i)--)
#define ft(i, s) for (auto i = s.begin(); i != s.end(); i++)
#define IOS ios::sync_with_stdio(0);cin.tie(0);
#define ff first
#define ss second
#define inp(a) f(i, a.size()) { cin >> a[i]; }
#define pb push_back
#define mp make_pair
#define MAX 1e9
#define endl '\n'
#define setb(x) __builtin_popcountll(x)
#define msb(x) (64 - __builtin_clzll(x))
#define rb(n) (n & (-n))
#define rbm(n) (n & (n - 1))
#define vec vector<int>
#define all(x) x.begin(), x.end()
#define sortd(x) sort(all(x), greater<int>());
#define sorta(x) sort(all(x));
#define mid (l + (r - l) / 2)
const int mod = 1e9+7;
int minvec(vec v) {int m = v[0];int ind = 0;f(i, v.size()){if (v[i] < m){m = v[i];ind = i;}}return ind;}
int maxvec(vec v) {int m = v[0];int ind = 0;f(i, v.size()){if (v[i] > m){m = v[i];ind = i;}}return ind;}
void yn(int val){if (val != 0){cout << "YES" << endl;}else{cout << "NO" << endl;}}
int cl(int x, int y){if (x % y == 0){return x / y;}else{return x / y + 1;}}
void pr(vec &x){f(i, x.size()){cout << x[i] << " ";}cout << endl;}
//cout << setprecision(15) << f << endl;
int gcd(int a, int b){if (b == 0)return a;a %= b;return gcd(b, a);}

int mpow(int x, int n){
  int ans = 1;
  while(n){
    if(n&1){
      ans=(ans*x)%mod;
    }
    x=(x*x)%mod;
    n=n>>1;
  }
  return ans;
}

int func(int n, int a, int b){
    if(n<1)
        return 0;
    if(n==1)
        return 1;
    if(n%a==0)
    return func(n/a,a,b) || func(n-b,a,b);
    else return func(n-b,a,b);
}

void solve()
{
    int n,a,b;
    cin>>n>>a>>b;
    if(a==1){
        yn((n-1)%b==0);
        return;
    }
    else if(b==0){
        yn(n%a==0);
        return;
    }
    int curr = 1;
    while(curr<=n){
        if((n-curr)%b==0){
            yn(1);
            return;
        }
        curr*=a;
    }
    yn(0);
}

signed main()
{
    IOS int t;
    // t=1;
    cin >> t;
    for (int loop = 0; loop < t; loop++)
        solve();
    return 0;
}