#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];

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
	set<ll>s;

	ll n, m; cin >> n >> m;
	for (int i = 1; i <= n; i++)f[i] = 1;
	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]++;
	}
	ll idx = 0;
	queue<ll>q;
	for (int i = 1; i <= n; i++) {
		if (din[i] == 0)q.push(i);
	}
	while (q.size()) {
		queue<ll>mid;
		ll pr = q.size();
		while (q.size()) {
			ll u = q.front(); q.pop();
			if (idx > n / 2)f[u] = 0;
			for (auto p = e[u].begin(); p != e[u].end(); p++) {
				ll v = *p; din[v]--;
				if (din[v] == 0) {
					mid.push(v);
				}
			}
		}
		idx += pr;
		while (mid.size()) {
			q.push(mid.front());
			mid.pop();
		}
	}
	for (int i = 1; i <= n; i++) {
		e[i].clear(); din[i] = 0;
	}
	if (idx != s.size()) {
		for (int i = 1; i <= n; i++)cout << 0;
		cout << endl;
		op.clear(); return;
	}
	//感觉有点多余，给出的应该是不分叉的线性关系
	//走一遍逆势
	for (auto p = op.begin(); p != op.end(); p++) {
		ll v = p->X, u = p->Y;
		e[u].push_back(v);
		din[v]++;
	}
	idx = 0;
	for (int i = 1; i <= n; i++) {
		if (din[i] == 0)q.push(i);
	}
	while (q.size()) {
		queue<ll>mid;
		ll pr = q.size();
		while (q.size()) {
			ll u = q.front(); q.pop();
			if (idx > n / 2)f[u] = 0;
			for (auto p = e[u].begin(); p != e[u].end(); p++) {
				ll v = *p; din[v]--;
				if (din[v] == 0) {
					mid.push(v);
				}
			}
		}
		idx += pr;
		while (mid.size()) {
			q.push(mid.front());
			mid.pop();
		}
	}
	for (int i = 1; i <= n; i++) {
		e[i].clear(); din[i] = 0;
	}
	for (int i = 1; i <= n; i++)cout << f[i];
	cout << endl;
	
	op.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;
}