It is always possible to rearrange the candies, unless candies of one specific color appears too often.

Let $$$x$$$ be the color with the highest number of candies. In case of ties, $$$x$$$ could be any of the candidates. Denote the number of candies with color $$$x$$$ as $$$a$$$.

Conclusion: The answer to the problem is YES if and only if there are no more than $$$n^2 - n$$$ candies of color $$$x$$$; — in other words, $$$a\le n^2 - n$$$ holds.

Proof of necessity: If there are more than $$$n^2 - n$$$ candies of color $$$x$$$, then there are less than $$$n$$$ candies not of color $$$x$$$. Since there are $$$n$$$ rows, at least one row will have no candies not of color $$$x$$$; therefore, such a row will be filled with candies of color $$$x$$$, violating the restrictions.

Proof of sufficiency: If there are no more than $$$n^2 - n$$$ candies of color $$$x$$$, we can always construct a solution with the method below.

First, if $$$a \lt n$$$ holds, then we can arrange the candies arbitrarily because no color has enough candies to fill a row or a column.

Otherwise, $$$n\le a\le n^2 - n$$$ holds.

We can put candies of color $$$x$$$ in $$$a_{1,1},a_{2,2},\ldots,a_{n,n}$$$. This will take $$$n$$$ candies of color $$$x$$$. Next, we choose $$$n$$$ candies not of color $$$x$$$ arbitrarily. We will put them at $$$a_{1,2},a_{2,3},\ldots,a_{n-1,n},a_{n,1}$$$. Lastly, we arrange other candies that have not yet been arranged arbitrarily. Since in the first step we put a candy of color $$$x$$$ in each row and column, and in the second step we put a candy not of color $$$x$$$ in each row and column, each row and column contains at least candies of two colors, satisfying the constraints.

#include<bits/stdc++.h>
#define F(i,x,y) for (int i=(x);i<=(y);i++)
#define R(i,x,y) for (int i=(x);i>=(y);i--)
#define p2i pair<int,int>
#define ll long long
#define fi first
#define se second
#define nb(x) (1<<((x)-1))
#define x1 __melody1
#define x2 __melody2
#define y1 __melody3
#define y2 __melody4
#define iosoptim ios::sync_with_stdio(0);cin.tie(0);
using namespace std;
 
const bool MT = 1;
int n,cnt[10005];
void ygg(){
    cin>>n;
    int tmp;
    bool flag=1;
    F(i,1,n*n) {
        cin>>tmp;
        ++cnt[tmp];
        if (cnt[tmp]>n*(n-1)) flag=0;
    }
    if (flag) cout<<"YES\n";
    else cout<<"NO\n";
}
void mark() {
    F(i,1,n*n) cnt[i]=0;
}
int main (){
#ifdef ONLINE_JUDGE
    iosoptim
#endif
    int testcases=1;
    if(MT) cin>>testcases;
    while(testcases--){
        ygg();
        mark();
    }
    return 0;
}