#include <bits/stdc++.h>
using namespace std;

#define SZ(a) (int)a.size()
#define ALL(a) a.begin(), a.end()
#define FOR(i, a, b) for (int i = (a); i<(b); ++i)
#define RFOR(i, b, a) for (int i = (b)-1; i>=(a); --i)
#define MP make_pair
#define PB push_back
#define F first
#define S second

typedef long long LL;
typedef pair<int, int> PII;
typedef vector<int> VI;

const int N = 400447;
const int mod[2] = {int(1e9) + 9, int(1e9) + 7};
const int p = 47;
VI g[N];
int pw[N][2];
int par[N];
bool used[N];
VI cycl;

int add(int a, int b, int md)
{
	a += b;
	if (a >= md)
		a -= md;
	return a;
}

int mult(int a, int b, int md)
{
	return LL(a) * b % md;
}

bool dfs1(int v, int from = -1)
{
	par[v] = from;
	used[v] = 1;
	bool ok = false;
	for (auto to : g[v])
	{
		if (to == from) continue;
		if (used[to])
		{
			cycl.PB(to);
			while (v != to)
			{
				cycl.PB(v);
				v = par[v];
			}			
			return true;
		}
		else
			ok |= dfs1(to, v);
		if (ok) return ok;
	}
	return false;
}
int sz[N];

PII dfs2(int v)
{
	sz[v] = 1;
	used[v] = 1;
	vector<PII> vec;
	for (auto to : g[v])
		if (!used[to]){
			vec.PB(dfs2(to));
			sz[v] += sz[to];
		}	
	sort(ALL(vec));
	PII hs = {(sz[v] + (sz[v] ^ 7)) , (sz[v] + (sz[v] ^ 7))};
	FOR (i, 0, SZ(vec)) hs = {	add(hs.first,  mult(pw[i + 1][0], vec[i].first,  mod[0]), mod[0]), 
								add(hs.second, mult(pw[i + 1][1], vec[i].second, mod[1]), mod[1])};
	return hs;
}

void solve()
{
	int n, m;
	cin >> n >> m;
	FOR (i, 0, m)
	{
		int a, b;
		cin >> a >> b;
		a--, b--;
		g[a].PB(b);
		g[b].PB(a);
	}
	if (m == n - 1)
	{
		cout << "YES\n";
	}
	else if (m > n)
	{
		cout << "NO\n";
	}
	else
	{
		dfs1(0, -1);
		FOR (i, 0, n) used[i] = 0;
		set<PII> s1, s2;
		for (auto v : cycl) used[v] = 1;
		FOR (i, 0, SZ(cycl)) 
		{
			if (SZ(cycl) % 2 == 0 && i % 2 == 0)
				s2.insert(dfs2(cycl[i]));
			else
				s1.insert(dfs2(cycl[i]));
		}
		cout << (SZ(s1) < 2 && SZ(s2) < 2 ? "YES\n" : "NO\n"); 
	}
	cycl.clear();
	FOR (i, 0, n)
	{
		g[i].clear();
		used[i] = 0;
	}
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	
	pw[0][0] = pw[0][1] = 1;
	FOR (i, 1, N) FOR (j, 0, 2) pw[i][j] = mult(pw[i - 1][j], p, mod[j]); 
	
	int t;
	cin >> t;
	while (t--)
	{
		solve();
	}
	
	return 0;
}