#define _CRT_SECURE_NO_WARNINGS 1
#include<bits/stdc++.h>
#include<unordered_map>
#include<unordered_set>
using namespace std;
#define ll long long
//#define double long double
#define lc u<<1
#define rc u<<1|1
#define X first
#define Y second
#define endl "\n"
#define int long long
const int N = 2e3 + 50;
const int M = 5e5 + 5;
const ll maxm = 1e18 + 5;
const ll mod = 998244353;
int dx[] = { -1,0,1,0 };
int dy[] = { 0,1,0,-1 };
//random_shuffle(id + 1, id + n + 1);
ll ksm(ll a, ll b);
ll gcd(ll a, ll b);

template<class T>inline void read(T& x) {
	x = 0;
	char c = getchar();
	while (!isdigit(c))c = getchar();
	while (isdigit(c))x = x * 10 + (c & 15), c = getchar();
}

void write(ll x)
{
	if (x < 0)
		putchar('-'), x = -x;
	if (x > 9)
		write(x / 10);
	putchar(x % 10 + '0');
	return;
}

bool f[N];
vector<pair<ll, ll> >op;
vector<vector<ll> >e(N);
int din[N];
set<ll>s;
ll n, m;

bool topo() {
	queue<ll>q;
	for (auto p = s.begin(); p != s.end(); p++) {
		if (din[*p] == 0)q.push(*p);
	}
	ll idx = 0;
	while (q.size()) {
		ll u = q.front(); q.pop();
		idx++;
		for (auto p = e[u].begin(); p != e[u].end(); p++) {
			ll v = *p; din[v]--;
			if (din[v] == 0) {
				q.push(v);
			}
		}
	}
	//cout << idx << " " << s.size() << endl;
	if (idx != s.size())return true;
	return false;
}

bool vis[N];
void bfs(ll s) {
	for (int i = 1; i <= n; i++)vis[i] = 0;
	queue<ll>q; q.push(s);
	ll idx = -1;
	while (q.size()) {
		ll u = q.front(); q.pop();
		if (vis[u])continue;
		idx++;
		vis[u] = 1;
		for (auto p = e[u].begin(); p != e[u].end(); p++) {
			ll v = *p;
			if (vis[v] == 0)q.push(v);
		}
	}
	//cout << s << " " << idx << endl;
	if (idx > n / 2)f[s] = 0;
}

void solve()
{
	//这个赛时就有思路，感觉是拓扑，正走一遍反走一遍，维护每个元素一定比他大的个数，一定比他小的个数
	//但是，怎么说，，先，怎么没思路
	//两遍拓扑即可，应该
	//感觉就是一个dfs,回来写，不对，这样被办法准确找到一定大于的元素，看看题再说
	//哦，不是，这纯水题啊，我还在想分叉怎么处理，结果根本不分叉，其实就直直一条线
	//建好图，入度为0 的点入队就完了，水的不行，裸的拓扑吧
	//我是罪人，m的边1到n^2不一定为连通图，所以判环应该要利用出现的节点的数量，不应该使用n

	//唉，还是缺少经验，写的太少，细节不到位
	//不光细节没处理好，裸拓扑不对，他给出的只是一种正确的可能，细节判断还得暴搜，拓扑判个环得了

	cin >> n >> m;
	for (int i = 1; i <= n; i++) {
		f[i] = 1; din[i] = 0;
	}
	for (int i = 1; i <= m; i++) {
		ll u, v; cin >> u >> v;
		s.insert(u); s.insert(v);
		op.push_back({ u,v });
	}
	//走一遍顺势
	for (auto p = op.begin(); p != op.end(); p++) {
		ll u = p->X, v = p->Y;
		e[u].push_back(v);
		din[v]++;
	}
	if (topo()) {
		for (int i = 1; i <= n; i++)cout << 0;
		cout << endl;
		op.clear(); s.clear();
		for (int i = 1; i <= n; i++) {
			e[i].clear();
		}
		return;
	}

	//当前为正序
	for (auto p = s.begin(); p != s.end(); p++) {
		if (f[*p])bfs(*p);
	}
	for (auto p = s.begin(); p != s.end(); p++) {
		e[*p].clear();
	}
	for (auto p = op.begin(); p != op.end(); p++) {
		ll v = p->X, u = p->Y;
		e[u].push_back(v);
	}
	for (auto p = s.begin(); p != s.end(); p++) {
		if (f[*p] == 1)bfs(*p);
	}
	for (int i = 1; i <= n; i++)cout << f[i];
	cout << endl;

	op.clear(); s.clear();
	for (int i = 1; i <= n; i++)e[i].clear();

	return;
}


signed main()
{
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
	int T = 1;
	cin >> T;
	//read(T);
	while (T--)
		solve();
	return 0;
}

/*PonyHex*/


ll ksm(ll a, ll b) {
	ll base = a;
	ll ans = 1;
	while (b) {
		if (b & 1)ans *= base % mod;
		ans %= mod;
		base *= base; base %= mod;
		b >>= 1;
	}
	return ans % mod;
}
ll gcd(ll a, ll b) {
	return b ? gcd(b, a % b) : a;
}