/*
60 + 0 + 100 + 64 = 224.
*/

#include <bits/stdc++.h>
using namespace std;
//#define int long long
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#define L(i, j, k) for (int i = (j); i <= (k); i++)
#define R(i, j, k) for (int i = (j); i >= (k); i--)
#define pb push_back
#define pii pair<int, int>
inline int read()
{
	int sum = 0, nega = 1;
	char ch = getchar();
	while (ch > '9'||ch < '0')
	{
	    if (ch == '-') nega = -1;
		ch = getchar();
	}
	while (ch <= '9' && ch >= '0') sum = sum * 10 + ch - '0', ch = getchar();
	return sum * nega;
}
const int N = (1 << 14) + 9, M = 509, mod = 998244353;
inline void add(int &x, int y) {x = (x + y) % mod;}
inline void del(int &x, int y) {x = (x - y + mod) % mod;}
int n, m, r, popc[N];
struct node
{
	int x, y;
};
vector<node> G[M], T[M];
int dp[N], f[16][16][N], to[N][16];
bool ok[16][16];
inline void solve(int x)
{
	L(i, 1, n)
		L(j, i, n)
		{
			bool fl = 0;
			for (auto v : T[x])
				if(v.x >= i && v.y <= j) fl = 1;
			for (auto v : G[x])
			{
				if(v.x >= i && v.x <= j && v.y > j) fl = 1;
				if(v.y >= i && v.y <= j && v.x < i) fl = 1;
			}
			ok[i][j] = fl;
		}
	memset(f, 0, sizeof(f));
	for (int i = 1; i < (1 << n); i *= 2)
        for (int j = 0; j < (1 << n); j += i * 2)
            for (int k = 0; k < i; k++)
                add(dp[j + k], dp[i + j + k]);
	L(j, 0, (1 << n) - 1) f[0][0][j] = dp[j];
	L(i, 0, n - 1)
		L(j, 0, (1 << n) - 1)
			if(popc[j] > i)
			{
				int p = to[j][i + 1], Ma = j ^ (1 << p), q = to[j][i + 2];
				if(!q) q = n;
				p++;
				L(k, 0, p - 1)
				{
					if(!f[i][k][j]) continue;
					if(k + 1 == p)
					{
						L(s, p, q)
							if(!ok[p][s]) add(f[i + 1][s][Ma | (1 << (s - 1))], f[i][k][j]);
					}
					else if(k < p && !ok[k + 1][p])
						add(f[i + 1][p][Ma | (1 << k)], f[i][k][j]);
				}
			}
	L(i, 0, (1 << n) - 1) dp[i] = f[popc[i]][n][i];
	return ;
}
signed main()
{
	n = read(), m = read(), r = read();
	L(i, 1, r)
	{
		int c = read(), x = read(), y = read(), dif = read();
		if(c == 1 && dif == 0) {puts("0"); return 0;}
		if(dif == 0) G[c].pb(node{x, y});
		else T[c].pb(node{x, y});
	}
	L(i, 0, (1 << n) - 1)
		L(j, 0, n - 1)
			if(i & (1 << j)) popc[i]++, to[i][popc[i]] = j;
	L(i, 0, (1 << n) - 1) dp[i] = 0;
	dp[(1 << n) - 1] = 1;
	L(i, 2, m) solve(i);
	int ans = 0;
	L(i, 0, (1 << n) - 1)
	{
		ans = (ans + dp[i]) % mod;
	}
	cout << ans << endl;
	return 0;
}