// #pragma GCC target("avx2")
// #pragma GCC optimize("O3")
// #pragma GCC optimize("unroll-loops")


#include<bits/stdc++.h>
using namespace std;
using uint = unsigned int;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
template<class T> using V = vector<T>;
template<class T> using VV = V<V<T>>;
template<class T> using VVV = V<VV<T>>;
template<class T> using VVVV = VV<VV<T>>;
#define rep(i,n) for(ll i=0ll;(i)<(n);(i)++)
#define REP(i,a,n) for(ll i=(a);(i)<(n);(i)++)
#define rrep(i,n) for(ll i=(n)-1;(i)>=(0ll);(i)--)
#define RREP(i,a,n) for(ll i=(n)-1;(i)>=(a);(i)--)
const long long INF = (1LL << 60);
const long long mod99 = 998244353;
const long long mod107 = 1000000007;
const long long mod = mod99;
#define eb emplace_back
#define be(v) (v).begin(),(v).end()
#define all(i,v) for(auto& (i) : (v))
#define UQ(v) sort(be(v)), (v).erase(unique(be(v)), (v).end())
#define UQ2(v,cmp) sort(be(v)), (v).erase(unique(be(v),cmp), (v).end())
#define UQ3(v,cmp) sort(be(v),cmp), (v).erase(unique(be(v)), (v).end())
#define UQ4(v,cmp,cmp2) sort(be(v), cmp), (v).erase(unique(be(v),cmp2), (v).end())
#define LB(x,v) (lower_bound(be(v),(x))-(v).begin())
#define LB2(x,v,cmp) (lower_bound(be(v),(x),(cmp))-(v).begin())
#define UB(x,v) (upper_bound(be(v),(x))-(v).begin())
#define UB2(x,v,cmp) (upper_bound(be(v),(x),(cmp))-(v).begin())
#define dout()  cout << fixed << setprecision(20)
#define randinit() srand((unsigned)time(NULL))

template<class T, class U> bool chmin(T& t, const U& u) { if (t > u){ t = u; return 1;} return 0; }
template<class T, class U> bool chmax(T& t, const U& u) { if (t < u){ t = u; return 1;} return 0; }


ll Rnd(ll L=0, ll R=mod99){return rand()%(R-L)+L;}



struct Combination{
    vector<long long> fac, inv, finv;
    long long MOD;
    Combination(long long N = 200100, long long p = 998244353) : fac(N, 1), inv(N, 1), finv(N, 1), MOD(p){
        for(long long i = 2; i < N; i++){
            fac[i] = fac[i-1] * i % MOD;
            inv[i] = MOD - inv[MOD%i] * (MOD/i) % MOD;
            finv[i] = finv[i-1] * inv[i] % MOD;
        }
    }
    long long com(long long n, long long k){
        if(n < k) return 0;
        if(n < 0 || k < 0) return 0;
        return fac[n] * finv[k] % MOD * finv[n-k] % MOD;
    }
    
    long long per(long long n, long long k){
        if(n < k) return 0;
        if(n < 0 || k < 0) return 0;
        return fac[n] * finv[n-k] % MOD;
    }
};

long long modpow(long long n, long long k, long long p = mod){
    long long a = n % p;
    long long ans = 1;
    while(k != 0) {
        if(k & 1) ans = ans * a % p;
        k /= 2;
        a = a * a % p;
    }
    
    return ans;  
}

// n^(-1) ≡ b (mod p) となる b を求める
long long modinv(long long n, long long p = mod) { 
//    if(n == 1) return 1;
//    return p - modinv(p % n) * (p / n) % p;
    return modpow(n, p - 2, p);
}

// n^k ≡ b (mod p) となる最小の k を求める
long long modlog(long long n, long long b, long long p = mod){
  
    long long sqrt_p = sqrt(p);
    unordered_map<long long , long long> n_pow;
    long long memo = 1;
    
    for(long long i = 0; i < sqrt_p; i ++){
        if(!n_pow.count(memo)) n_pow[memo] = i;
        memo = memo * n % p; 
    }
    
    memo = modinv(memo, p);
    long long ans = 0;
    while(!n_pow.count(b)){
        if(ans >= p) return -1;
        ans += sqrt_p;
        b = b * memo % p;
    }
  
    ans += n_pow[b];
    return ans % (p - 1);

}

// ax + by = gcd(a, b) を満たす (x, y) が格納される
long long ext_gcd(long long a, long long b, long long &x, long long &y){
    if(b == 0){
        x = 1;
        y = 0;
        return a;
    }
    long long d = ext_gcd(b, a%b, y, x);
    y -= a/b*x;
    return d;
}

void solve(){
    ll a,b,n;
    cin >> a >> b >> n;
    ll nn = n;
    ll cnt = 0;
    ll xx,yy;
    ext_gcd(a, b, xx, yy);

    while(nn > b){
        ll x = xx, y = yy;
        y %= a;
        y *= (nn%a);
        y %= a;
        if(y > 0) y -= a;
        
        ll nx = nn - y*b;
        // cout << nn << " " << nx << endl;
        if(cnt){
            if(nn <= n and n <= nx and (n-nn)%b == 0){
                cout << "Yes\n";
                return;
            }
        }
        cnt++;
        nn = (nn - y*b) / a;
    }
    
    ll x = xx, y = yy;
    x %= b;
    x *= (nn % b);
    x %= b;
    if(x <= 0) x += b;
    // cout << nn << " " << x << " " << n << " " << x*a << endl;
    if(n > x*a){
        cout << "No\n";
        return;
    }
    cout << "Yes\n";




}





int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    int t=1;
    cin >> t;
    rep(i,t) solve();
}