#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> Vll;
typedef vector<pair<int, int> > vpii;
typedef vector<pair<ll, ll> > vpll;

const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const double pi = acos(-1.0);

inline ll qmi (ll a, ll b) {
ll ans = 1;
while (b) {
if (b & 1) ans = ans * a%mod;
a = a * a %mod;
b >>= 1;
}
return ans;
}
inline int read () {
int x = 0, f = 0;
char ch = getchar();
while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
return f?-x:x;
}
template<typename T> void print(T x) {
if (x < 0) putchar('-'), x = -x;
if (x >= 10) print(x/10);
putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
return (a + b) %mod;
}
inline ll inv (ll a) {
return qmi(a, mod - 2);
}

void solve() {
int n;
cin >> n;
vector<int> a(n);
map<int, int> cnt;
int maxv = 0;
for (int i = 0; i < n; i ++){
cin >> a[i];
maxv= max(maxv, a[i]);
}
int mm = maxv;
map<int, int> b;
for (int i = 2; i <= mm / i; i ++) {
if (mm % i !=0) continue;
int c = 0;
while (mm % i == 0) {
mm /= i;
c ++;
}
b[i] = c;
}
if (mm > 1) {
b[mm] ++;
}
for (auto t : a) {
map<int, int> yinzi;
for (int j = 2 ;j <= t / j; j ++) {
if (t % j != 0) continue;
int c = 0;
while (t % j == 0) {
t /= j;
c ++;
}
yinzi[j] = c;
}
if (t > 1) {
yinzi[t]++;
}
for (auto [x, cnt] : yinzi) {
if (cnt != b[x] && x != 2 && x != 3) {
puts("NO");
return;
}
}
for (auto [x, cnt] : b) {
if (cnt != yinzi[x] && x != 2 && x != 3) {
puts("NO");
return;
}
}
}
puts("YES");
}
int main () {
// ios::sync_with_stdio(0),cin.tie(0), cout.tie(0);
int t;
t =1;
cin >> t;
while (t --) solve();
return 0;
}

将2 和3的因子全部去掉,比较大小,看是否相同不相同则no,否则yes