/* Kasioopea 2014
 * http://ksp.mff.cuni.cz/akce/kasiopea/
 *
 * Autorské řešení úlohy 26-K-6: Zajímavá posloupnost
 *
 */
#include <algorithm>
#include <iostream>

using namespace std;

const int MAXN = 200000;

int n, a[MAXN], values[MAXN];
int prev[MAXN], next[MAXN], last[MAXN];

bool je_zajimava(int beg, int end) {
	if (beg == end) return false;
	for (int i = 0; i < (end - beg); ++i) {
		int mid = ((i & 1) ? beg + i / 2 : end - 1 - i / 2);
		if (prev[mid] < beg && next[mid] >= end)
			return je_zajimava(beg, mid) || je_zajimava(mid+1, end);
	}
	return true;
}

int main() {
	ios_base::sync_with_stdio(false);
	int T;
	cin >> T;
	while (T--) {
		cin >> n;
		for (int i = 0; i < n; ++i) {
			cin >> a[i];
			values[i] = a[i];
		}
		sort(values, values + n);
		int m = unique(values, values + n) - values;
		for (int i = 0; i < n; ++i)
			a[i] = lower_bound(values, values + m, a[i]) - values;

		for (int i = 0; i < m; ++i) last[i] = -1;
		for (int i = 0; i < n; ++i) {
			prev[i] = last[a[i]];
			last[a[i]] = i;
		}
		for (int i = 0; i < m; ++i) last[i] = n;
		for (int i = n-1; i >= 0; --i) {
			next[i] = last[a[i]];
			last[a[i]] = i;
		}

		cout << (je_zajimava(0, n) ? "nezajimava\n" : "zajimava\n");
	 }
	return 0;
}