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 <iostream>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
#include <cassert>
#include <iomanip>

using namespace std;

typedef long long ll;
typedef long double ld;

#define int ll
#define mp make_pair
#define pb push_back
#define pi pair<int, int>
#define vi vector<int>
#define vpi vector<pi>
#define vvpi vector<vpi>
#define vs vector<string>
#define ff first
#define ss second
#define vvi vector<vi>
#define len(v) int(v.size())
#define all(v) v.begin(), v.end()
#define in_rng(i, n) i >= n ? i - n : ((i >= 0) ? i : i + n)


ll bin_pow(ll base, ll deg, ll md) {
    if (!deg)
        return 1;
    if (deg % 2)
        return base * bin_pow(base, deg - 1, md) % md;
    ll res = bin_pow(base, deg / 2, md);
    return res * res % md;
}

//const int max_a = 3e5 + 7;
template <typename T1, typename T2>
bool cmin(T1& a, const T2& b) {
    if (a <= b)
        return false;
    a = b;
    return true;
}

template <typename T1, typename T2>
bool cmax(T1& a, const T2& b) {
    if (a >= b)
        return false;
    a = b;
    return true;
}

int get_rand_int(int mn, int mx) {
    return rand() % (mx - mn + 1) + mn;
}


const int md = 1e9 + 7;

int inv(int a) {
    return bin_pow(a, md - 2, md);
}

int gcd(int a, int b) {
    if (!b)
        return a;
    return gcd(b, a % b);
}

const int max_n = 1e6 + 7;


int get_bit(int m, int index) {
    return (bool)(m >> index) & 1;
}

int invert(int num, int n) {
    for (int i = 0; i <= n; ++i) {
        num ^= 1 << i;
    }
    return num;
}

void run() {
    int n, a, b;
    cin >> n >> a >> b;
    if (a == 1) {
        if ((n - 1) % b == 0)
            cout << "Yes\n";
        else
            cout << "No\n";
        return;
    }
    int mult = 1;
    for (; mult <= n; mult *= a) {
        int res = mult;
        if ((n - mult) % b == 0) {
            cout << "Yes\n";
            return;
        }

    }
    cout << "No\n";
}


signed main() {
    cin.tie(0);
    cout.tie(0);
    ios_base::sync_with_stdio(0);
    /*
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    //*/
    //cout << fixed << setprecision(12);
    size_t t = 1;
    std::cin >> t;
    for (size_t i = 0; i < t; ++i) {
        //arr.clear();
        run();
    }
    int tmp;
    //std::cin >> tmp;
}