A. XXXXX

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

An array aa is a subarray of an array bb if aa can be obtained from bb 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 tt (1≤t≤5)(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 nn and xx (1≤n≤1051≤n≤105, 1≤x≤1041≤x≤104) — the number of elements in the array aa and the number that Ehab hates.

The second line contains nn space-separated integers a1a1, a2a2, ……, anan (0≤ai≤1040≤ai≤104) — the elements of the array aa.

Output

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

Example

input

Copy

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

output

Copy

2
3
-1

Note

In the first test case, the subarray [2,3][2,3] has sum of elements 55, which isn't divisible by 33.

In the second test case, the sum of elements of the whole array is 66, which isn't divisible by 44.

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

解題說明：此題是一道數學題，爲了確保不被y整除，先按餘數進行求和，所有的數都是b的倍數，那無論a的那個子區間的和都是b的倍數。a的總和不是b的倍數，那答案就是整個數組的長度。a的總和是b的倍數，則從左往右找第一個不是b的倍數的數的下標x，從右往左找第一個不是b的倍數的數的下標y。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>

using namespace std;

int main() 
{
	int t, a[100000], b[100000], y, i, r, l, n;
	scanf("%d", &t);
	while (t--) 
	{
		scanf("%d", &n);
		scanf("%d", &y);
		for (i = 1; i <= n; i++) 
		{
			scanf("%d", &a[i]);
			a[i] = a[i] % y;
			b[i] = a[i];
		}
		a[0] = 0;
		for (i = 1; i <= n; i++) 
		{
			a[i] += a[i - 1];
		}
		if (a[n] % y != 0) 
		{
			printf("%d\n", n);
		}
		else 
		{
			r = n;
			while (r >= 1 && b[r] % y == 0) 
			{
				r--;
			}
			r--;
			l = 1;
			while (l <= n && b[l] % y == 0) 
			{
				l++;
			}
			if (n - l > r) 
			{
				printf("%d\n", n - l);
			}
			else 
			{
				printf("%d\n", r);
			}
		}
	}
	return 0;
}