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>
/*#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;*/
using namespace std;

typedef pair<int,int> pi;
typedef pair<int,pair<int,int>> pii;
typedef vector<int> vi;
typedef vector<pi> vpi;
typedef vector<pii> vpii;
typedef vector<vi> vvi;
//typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> pbds;
// order_of_key (val): returns the no. of values less than val
// find_by_order (k): returns the iterator to kth largest element.(0-based)
#define w(x) int x; cin>>x; while(x--)
#define fo(i,a,b) for(int i=a;i<b;i++)
#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define pql priority_queue<int>
#define pqs priority_queue<int,vi,greater<int> >
#define all(x) x.begin(), x.end()
#define fill(x) memset(x, -1, sizeof(x))
#define fill0(x) memset(x, 0, sizeof(x))
#define sortall(x) sort(all(x))
#define gc getchar_unlocked
#define tr(it, a) for(auto it = a.begin(); it != a.end(); it++)
#define ps(x,y) fixed<<setprecision(y)<<x
#define setbits(x) __builtin_popcountll(x)
#define PI 3.1415926535897932384626
#define inf 1e18
#define int long long int
#define mod 1000000007
#define mod1 4294967296  // 2^32

int mpow(int base, int exp) {
    base %= mod;
    int result = 1;
    while (exp > 0) {
        if (exp & 1) result = (result * base) % mod;
        base = (base * base) % mod;
        exp >>= 1;
    }
    return result;
}


void solve() {
    int n,a,b;
    cin>>n>>a>>b;
    if(a==1){
        if((n-1)%b==0){
            cout<<"Yes"<<endl;
        } else {
            cout<<"No"<<endl;
        }
        return;
    }
    int t=1;
    while(n>=t){
        int x=n-t;
        if(x%b == 0){
            cout<<"Yes"<<endl;
            return;
        }
        t*=a;
    }
    cout<<"No"<<endl;
}


int32_t main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    // int i=1;
    w(x)
    {
      // cout<<"Case #"<<i<<": ";
        solve();
      // i++;
    }
    // cerr<<"Time : "<<1000*((double)clock())/(double)CLOCKS_PER_SEC<<"ms"<<endl;
    return 0;
}