#include <bits/stdc++.h>
// #include "stdafx.h"
// #pragma warning(disable : 4996) //_CRT_SECURE_NO_WARNINGS
using namespace std;
#define gc getchar_unlocked
#define fo(i,n) for(i=0;i<n;i++)
#define Fo(i,k,n) for(i=k;k<n?i<n:i>n;k<n?i+=1:i-=1)
#define ll long long
#define si(x)	scanf("%d",&x)
#define sl(x)	scanf("%lld",&x)
#define ss(s)	scanf("%s",s)
#define pi(x)	printf("%d\n",x)
#define pl(x)	printf("%lld\n",x)
#define ps(s)	printf("%s\n",s)
#define deb(x) cout << #x << "=" << x << endl
#define deb2(x, y) cout << #x << "=" << x << "," << #y << "=" << y << endl
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define all(x) x.begin(), x.end()
#define clr(x) memset(x, 0, sizeof(x))
#define sortall(x) sort(all(x))
#define tr(it, a) for(auto it = a.begin(); it != a.end(); it++)
#define PI 3.1415926535897932384626
typedef pair<int, int>	pii;
typedef pair<ll, ll>	pl;
typedef vector<int>		vi;
typedef vector<ll>		vl;
typedef vector<pii>		vpii;
typedef vector<pl>		vpl;
typedef vector<vi>		vvi;
typedef vector<vl>		vvl;
int mpow(int base, int exp); 
void ipgraph(int m);
void dfs(int u, int par);
const int mod = 998244353;
const int N = 3e5, M = N;
//=======================

vi g[N];
int a[N], col[N];
int n, m, valid;

void color(int u, int p, int c)
{
	col[u] = c;
	int nc = 3-c;
	for(int v: g[u])
	{
		if(col[v]){
			if(col[v] != nc) valid = false;
		}
		else color(v, u, 3-c);
	}
}

void solve()
{
	int i, x, y;
	cin >> n >> m;
	fo(i, n) g[i].clear(), col[i] = 0;
	valid = true;
	
	while(m--)
	{
		cin >> x >> y; x--, y--;
		g[x].pb(y);
		g[y].pb(x);
	}
	//This won't get AC in the case 
	//where the graph will be disconnected
	//but the overall idea is this...
	color(0, -1, 1);
	if(!valid) cout << 0 << endl;
	else {
		ll ans = 0;
		int cnt = 0;
		fo(i, n) cnt += col[i] == 1;
		ans = (mpow(2, cnt) + mpow(2, n-cnt)) % mod;
		cout << ans << endl;
	}
}

int main()
{
	int t;
	//freopen("input.txt", "r", stdin);
	//freopen("output.txt", "w", stdout);
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	int i,n,k,j;
	cin >> t;
	while(t--) solve();
	return 0;
} 

int mpow(int base, int exp) {
	base %= mod;
	int result = 1;
	while (exp > 0) {
	if (exp & 1) result = ((ll)result * base) % mod;
	base = ((ll)base * base) % mod;
	exp >>= 1;
	}
	return result;
}