#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll N=1e5+5; 
const ll M=999999998;
//快速幂
ll quick(ll x,ll y){    
	ll z=1;
	while(y!=0){
		if(y%2==1){
			z=(z*x)%M;
		}
		x=(x*x)%M;
		y=y/2;
	}
	return z%M;//z%M即为2的t2或3的t3次方取与的结果了 
}
int main()
{
	int T;
	scanf("%d",&T);
	while(T--){
		int n,m;
		scanf("%d %d",&n,&m);
		int a2[N],a3[N];
        int b2[N],b3[N];
        memset(a2,0,sizeof(a2));
        memset(a3,0,sizeof(a3)); 
        memset(b2,0,sizeof(b2));
        memset(b3,0,sizeof(b3));//不知道这个函数的可自行查阅，在这里只是起一个初始化的作用
		while(m--){    
			ll l,r,x;
			scanf("%lld %lld %lld",&l,&r,&x);
			if(x==2){
				a2[l]++;
				a2[r+1]--;
			}else{
				a3[l]++;
				a3[r+1]--;
			}
		}
	    int t2=a2[1],t3=a3[1];
	    for(ll i=1;i<=n;i++){   //差分，本题重点
		    b2[i]=b2[i-1]+a2[i];
		    t2=min(t2,b2[i]);
		    b3[i]=b3[i-1]+a3[i];
		    t3=min(t3,b3[i]);
	    }
	    ll c=quick(2,t2)*quick(3,t3)%M;   
	    printf("%lld\n",c);
    }
	return 0;
}