//https://www.infoarena.ro/problema/perle

//#pragma GCC optimize("O3")   
//#pragma GCC optimize("Ofast") 
//#pragma GCC optimize("fast-math") 
//#pragma GCC optimize("unroll-loops") 
//#pragma GCC optimize("inline")  
//#define _USE_MATH_DEFINES
//#define _CRT_SECURE_NO_WARNINGS

#include <iostream>
#include <fstream>
//#include <vector>
//#include <cstring>
//#include <cmath>
//#include <bitset>
#include <queue>
//#include <stack>
//#include <utility>
//#include <algorithm>
//#include <string>
//#include <map>
//#include <unordered_map>
//#include <set>
//#include <unordered_set>
//#include <cstdint>
//#include <climits>
//#include <iomanip>
//#include <cstdio>
//#include <tuple>

using namespace std;

ifstream fin("perle.in");
ofstream fout("perle.out");

string v;

bool verifA(int st, int dr); // A
bool verifB(int st, int dr); // B
bool verifC(int st, int dr); // C

bool verifA(int st, int dr) // A (nu cred ca am nevoie da il las)
{
	return (dr == st && (v[st] == '1' || v[st] == '2' || v[st] == '3'));
}

bool verifB(int st, int dr) // C
{
	if (dr - st >= 1 && v[st] == '2') // are mai mult de 2 elemente -> 2B
		return verifB(st + 1, dr);

	if (dr - st >= 2 && v[st] == '1' && v[st + 2] == '3') // are mai mult de 5 element -> pun 1A3AC
		return (verifA(st + 1, st + 1) && verifA(st + 3, st + 3) && verifC(st + 4, dr));

	return false; // nu merge B :(
}

bool verifC(int st, int dr) // C
{
	if (st == dr && v[st] == '2') // are un element -> 2
		return true;

	if (dr - st == 2 && v[st] == '1' && v[st + 1] == '2') // are 3 elemente -> pun 12A
		return verifA(st + 2, dr);

	if (dr - st >= 2 && v[st] == '3')
		for (int i = st + 1; i < dr; ++i) // aici e complicat. daca are mai mult de 3 elemente as vrea sa gasesc o combinatie de un B si un C care sa dea
			if (verifB(st + 1, i) && verifC(i + 1, dr))
				return true;


	return false; // nu merge C :(
}

int main()
{
	//ios_base::sync_with_stdio(false);
	//cin.tie(nullptr);
	//cout.tie(nullptr);

	int n, nr, i;

	fin >> n;

	// trebuie o funcitie cre ia un interval st dr din tot vectorul si verifica daca exista ceva care sa intre acolo

	while (n--)
	{
		fin >> nr;

		v.clear();
		for (i = 1; i <= nr; ++i)
		{
			int x;
			fin >> x;
			v += ('0' + x);
		}

		fout << (verifA(0, (int)v.size() - 1) || verifB(0, (int)v.size() - 1) || verifC(0, (int)v.size() - 1)) << "\n";
		// sa vezi ca trebe mai eficient ...
	}
	
	return 0;
}