Problem: Ehab loves number theory, but for some reason he hates the number x. Given an array a, find the length of its longest subarray such that the sum of its elements isn't divisible by x, or determine that such subarray doesn't exist.

An array a is a subarray of an array b if a can be obtained from b by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.

Input

The first line contains an integer t (1 ≤ t ≤ 5) — the number of test cases you need to solve. The description of the test cases follows.

The first line of each test case contains 2 integers n and x (1 ≤ n ≤ 10^5, 1 ≤ x ≤ 10^4) — the number of elements in the array a and the number that Ehab hates.

The second line contains n space-separated integers a_1, a_2, …, a_{n} (0 ≤ a_i ≤ 10^4) — the elements of the array a.

Output

For each testcase, print the length of the longest subarray whose sum isn't divisible by x. If there's no such subarray, print -1.

Example

Input


3
3 3
1 2 3
3 4
1 2 3
2 2
0 6


Output


2
3
-1

Note

In the first test case, the subarray [2,3] has sum of elements 5, which isn't divisible by 3.

In the second test case, the sum of elements of the whole array is 6, which isn't divisible by 4.

In the third test case, all subarrays have an even sum, so the answer is -1.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int cansel_sync = (ios::sync_with_stdio(0), cin.tie(0), 0);
const int N = 200010;
const int inf = ~0u >> 2;
const long long INF = ~0ull >> 2;
long long read() {
  long long x;
  if (scanf("%lld", &x) == -1) exit(0);
  return x;
}
const double pi = acos(-1.0);
double readf() {
  double x;
  if (scanf("%lf", &x) == -1) exit(0);
  return x;
}
template <typename T>
T sqr(const T &x) {
  return x * x;
}
int mod = (1 ? 1000000007 : 998244353);
long long mul(long long a, long long b, long long m = mod) { return a * b % m; }
long long qpow(long long a, long long b, long long m = mod) {
  long long ans = 1;
  for (; b; a = mul(a, a, m), b >>= 1)
    if (b & 1) ans = mul(ans, a, m);
  return ans;
}
long long getinv(long long v, long long m = mod) { return qpow(v, m - 2, m); }
long long n, x, t, l, r, sum;
void solve() {
  cin >> n >> x;
  sum = 0;
  l = inf, r = -inf;
  for (long long i = (0), _ = (n); i < _; i++) {
    cin >> t;
    if (t % x != 0) l = min(l, i), r = max(r, i);
    sum += t % x;
  }
  if (sum % x != 0) {
    cout << n << endl;
    return;
  }
  if (l != inf) {
    cout << max(n - l - 1, r) << endl;
    return;
  }
  cout << -1 << endl;
}
signed main() {
  long long T;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}