#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define op(x) ((x&1)?x+1:x-1)
#define odd(x) (x&1)
#define even(x) (!odd(x))
#define lc(x) (x<<1)
#define rc(x) (lc(x)|1)
#define lowbit(x) (x&-x)
#define mp(x,y) make_pair(x,y)
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
using namespace std;
const int MAXN = 14,MAXM = 510,mod = 998244353;
void add(ll& x,ll y){x = (x+y)%mod;}

int n,m,r;

typedef array<int,3> tr;
typedef array<int,15> node;
vector<tr>o[MAXM];

ll dp[2][1<<MAXN],f[1<<MAXN],ans;

int col[MAXN],nxt[MAXN],L[MAXN],R[MAXN];
ll s;

int tot;
vector<node>tmp;
vector<int>to[1<<MAXN];

void dfs(int r){
	if(r==n){ 
		//确定每个刷子的位置
		int t=0;
		rep(i,0,n-1)R[col[i]]=i;
		per(i,n-1,0)L[col[i]]=i;
		rep(i,0,n-1)if(s>>i&1){
			if(L[i]<i)t|=(1<<L[i]);
			else t|=(1<<R[i]);
		}

		node T;rep(i,0,n-1)T[i] = col[i]; T[n] = t;
		tot = tmp.size();tmp.push_back(T);
		to[s].push_back(tot);
		return;
	}
	if(s>>r&1)col[r]=r,dfs(r+1);
	else{
		if(r){
			int f=1;
			if(s>>(r-1)&1){
				if(r>1 && col[r-2]==col[r-1])f=0;
			}
			if(f){
				col[r]=col[r-1];
				dfs(r+1);
			}
		}

		if( (nxt[r]^n) ){
			int f=1;
			if(r && col[r-1] == nxt[r])f = 0;
			if(f){
				col[r]=nxt[r];
				dfs(r+1);
			}
		}
	}
}

void trans(int c){
	int p=(c+1)&1;
	memset(f,0,sizeof f);
	rep(S,1,(1<<n)-1)add(f[S],dp[p^1][S]),dp[p^1][S]=0;
	rep(i,0,n-1)rep(S,1,(1<<n)-1)if(S>>i&1)add(f[S^(1<<i)],f[S]);
	rep(S,1,(1<<n)-1)if(f[S]){
		ll val=f[S];
		for(auto id:to[S]){
			node& col=tmp[id];
			int cur=-1;
			int fl=1;
			for(auto [x,y,d]:o[c+1]){
				if((col[x]!=col[y])!=d){fl=0;break;}
			}
			if(!fl)continue;
			
			add(dp[p][col[n]],val);
		}
	}	
}

int main(){
	cin>>n>>m>>r;
	rep(i,1,r){
		int c,x,y,d;cin>>c>>x>>y>>d;
		x--,y--;
		if(c==1){
			if(!d)return (void)(cout<<"0\n"),0;
		}else o[c].push_back({x,y,d});
	}

	rep(i,1,(1<<n)-1){
		s=i;
		nxt[n] = n;
		per(i,n-1,0)nxt[i]=(s>>i&1)?(i):(nxt[i+1]);
		dfs(0);
	}

	dp[1][(1<<n)-1] = 1;
	rep(i,1,m-1)trans(i);

	rep(S,0,(1<<n)-1)if(dp[m&1][S])add(ans,dp[m&1][S]);
	cout<<ans<<endl;
    return 0;
}