cf round 83 c

Suppose you are performing the following algorithm. There is an array v1,v2,…,vn filled with zeroes at start. The following operation is applied to the array several times — at i-th step (0-indexed) you can:

either choose position pos (1≤pos≤n) and increase vpos by ki;
or not choose any position and skip this step.
You can choose how the algorithm would behave on each step and when to stop it. The question is: can you make array v equal to the given array a (vj=aj for each j) after some step?

Input
The first line contains one integer T (1≤T≤1000) — the number of test cases. Next 2T lines contain test cases — two lines per test case.

The first line of each test case contains two integers n and k (1≤n≤30, 2≤k≤100) — the size of arrays v and a and value k used in the algorithm.

The second line contains n integers a1,a2,…,an (0≤ai≤1016) — the array you’d like to achieve.

Output
For each test case print YES (case insensitive) if you can achieve the array a after some step or NO (case insensitive) otherwise.

input Copy
5
4 100
0 0 0 0
1 2
1
3 4
1 4 1
3 2
0 1 3
3 9
0 59049 810

output Copy
YES
YES
NO
NO
YES

int t, n,k;
ll a[55];
int index, ji[64];

int main() {
	scanf("%d", &t);
	while (t--) {
		scanf("%d%d", &n,&k);
		for (int i = 0; i < n; i++)
			scanf("%lld", a + i);
		

		ms(ji, 0);
		bool flag = true;
		for (int i = 0; i < n; i++) {
			index = 0;
			if (!flag) break;

			while (a[i]) {
				if ((a[i] - 1) % k == 0 && ji[index] == 0) {
					a[i]--;
					if(a[i]) a[i] /= k;
					ji[index++] = 1;
				}
				else if (a[i] % k == 0) {
					a[i] /= k;
					index++;
				}
				else {
					flag = false;
					break;
				}
			}
		}

		if (flag)
			printf("YES\n");
		else
			printf("NO\n");
	}

	return 0;
}