from sys import stdin, stdout


def read_int():
    return int(stdin.readline().strip())


t = read_int()
for case_num in range(t):
    n = read_int()
    counter = [0 for _ in range(26)]

    for i in range(n):
        s = stdin.readline().strip()
        for c in s:
            counter[ord(c) - ord('a')] += 1
    ok = True
    for count in counter:
        if count % n != 0:
            ok = False
            break
    stdout.write('YES\n' if ok else 'NO\n')

from sys import stdin, stdout


def read_int():
    return int(stdin.readline().strip())


def read_ints():
    return map(int, stdin.readline().strip().split(' '))


n = read_int()
a = list(read_ints())
a.sort()
s = sum(a)
cost = s - n
j = 2
while pow(j, n - 1) <= s * 2:
    b = [1]
    for k in range(n - 1):
        b.append(b[-1] * j)
    tot = sum([abs(a[i] - b[i]) for i in range(n)])
    cost = min(cost, tot)
    j += 1
print(cost)

#include <cstdio>
#include <iostream>
#include <vector>

using namespace std;
typedef long long ll;

template <typename T> void read(T &x) {
  x = 0;
  char c = getchar();
  T sig = 1;
  for (; !isdigit(c); c = getchar())
    if (c == '-')
      sig = -1;
  for (; isdigit(c); c = getchar())
    x = (x << 3) + (x << 1) + c - '0';
  x *= sig;
}

class Solution {
public:
  void solve() {
    int n;
    read(n);
    vector<ll> a(n);
    for (int i = 0; i < n; ++i)
      read(a[i]);
    if (n == 1) {
      printf("1 1\n%lld\n1 1\n1\n1 1\n-1\n", -a[0]);
      return;
    }
    printf("1 %d\n", n - 1);
    for (int i = 0; i < n - 1; ++i) {
      ll r = (a[i] % n + n) % n;
      a[i] += r * (n - 1);
      printf("%lld ", r * (n - 1));
    }
    printf("\n2 %d\n", n);
    for (int i = 1; i < n; ++i) {
      ll r = (a[i] % n + n) % n;
      a[i] += r * (n - 1);
      printf("%lld ", r * (n - 1));
    }
    printf("\n1 %d\n", n);
    for (ll num : a)
      printf("%lld ", -num);
  }
};

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  Solution solution = Solution();
  solution.solve();
}

#include <cstdio>
#include <iostream>
#include <vector>

using namespace std;

template <typename T> void read(T &x) {
  x = 0;
  char c = getchar();
  T sig = 1;
  for (; !isdigit(c); c = getchar())
    if (c == '-')
      sig = -1;
  for (; isdigit(c); c = getchar())
    x = (x << 3) + (x << 1) + c - '0';
  x *= sig;
}

class Solution {
public:
  void solve() {
    int n;
    read(n);
    vector<int> a(n);
    int sum = 0, hi = 0;
    for (int i = 0; i < n; ++i) {
      read(a[i]);
      sum += a[i];
      hi = max(hi, a[i]);
    }
    if (hi > sum - hi) {
      printf("T\n");
      return;
    }
    printf(sum % 2 == 0 ? "HL\n" : "T\n");
  }
};

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t;
  read(t);
  while (t--) {
    Solution solution = Solution();
    solution.solve();
  }
}

Consider route $$$i-3\rightarrow i-2\rightarrow i-1\rightarrow i\rightarrow i-1\rightarrow i-2\rightarrow i-1\rightarrow i$$$.

It can be rearranged to $$$i-3\rightarrow i-2\rightarrow i-1\rightarrow i-2\rightarrow i-1\rightarrow i\rightarrow i-1\rightarrow i$$$, and the length does not change. Moreover, both routes pass level $$$i-2$$$, $$$i-1$$$ and $$$i$$$ at least twice. So these two routes come to the same optimal value. The second part of the rearranged route is just what we have considered ($$$i-2\rightarrow i-1\rightarrow i\rightarrow i-1\rightarrow i$$$), yet the first part of the rearranged route ($$$i-3\rightarrow i-2\rightarrow i-1\rightarrow i-2$$$) is no better than $$$L[i-2]$$$ so we do not need to consider it.

#include <cstdio>
#include <iostream>
#include <vector>

using namespace std;
typedef long long ll;

template <typename T> void read(T &x) {
  x = 0;
  char c = getchar();
  T sig = 1;
  for (; !isdigit(c); c = getchar())
    if (c == '-')
      sig = -1;
  for (; isdigit(c); c = getchar())
    x = (x << 3) + (x << 1) + c &mdash; '0';
  x *= sig;
}

class Solution {
  struct Node {
    ll time = 0;
    int idx = 0, left = 0;
    bool operator<(const Node &other) const { return time > other.time; }
  };

public:
  void solve() {
    ll n, r1, r2, r3, d;
    read(n), read(r1), read(r2), read(r3), read(d);
    vector<ll> a(n + 1), kill_all(n + 1), leave_one(n + 1);
    for (int i = 1; i <= n; ++i) {
      read(a[i]);
      kill_all[i] = r1 * a[i] + r3;
      leave_one[i] = min(r2, r1 * (a[i] + 1)) + r1;
    }
    vector<ll> L(n + 2), R(n + 2);
    R[n] = min(kill_all[n], leave_one[n] + d * 2);
    for (int i = n &mdash; 1; i >= 1; --i)
      R[i] = R[i + 1] + d * 2 + min(kill_all[i], leave_one[i]);
    ll cost = R[1];
    L[0] = R[n + 1] = -d;
    for (int i = 1; i <= n; ++i) {
      L[i] = L[i &mdash; 1] + d + min(kill_all[i], leave_one[i] + d * 2);
      if (i >= 2)
        L[i] = min(L[i], L[i &mdash; 2] + d * 4 +
                             min(kill_all[i &mdash; 1], leave_one[i &mdash; 1]) +
                             min(kill_all[i], leave_one[i]));
      cost = min(cost, L[i] + R[i + 1] + d);
    }
    cout << cost;
  }
};

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  Solution solution = Solution();
  solution.solve();
}