#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
using vi = vector<int>;
using vl = vector<ll>;
const int MOD = 1e9+7;
#define F first
#define S second
#define PB push_back
#define all(a) (a).begin(), (a).end()
#define rall(a) (a).rbegin(), (a).rend()
struct Info {
int sgn, x, y;
};
ll n, m;
int k;
ll mod_pow(ll x, ll y) {
ll ans = 1;
while(y) {
if(y & 1) ans = ans*x%MOD;
x = x*x%MOD;
y >>= 1;
}
return ans;
}
int main() {
ios::sync_with_stdio(0); cin.tie(0);
cin >> n >> m >> k;
vector<Info> A(k);
for(int i = 0; i < k; i++) {
char c; cin >> c >> A[i].y >> A[i].x;
A[i].sgn = c == '+';
}
// answer = |alternating rows| + |alternating cols| - |alternating both|
ll ans = 0;
// rows (group by y check all x)
map<int, int> total, satisfy;
bool ok = true;
for(int i = 0; i < k; i++) {
total[A[i].y]++;
satisfy[A[i].y] += (A[i].x % 2 == A[i].sgn);
}
for(auto p : total) {
if(!(p.S == satisfy[p.F] || !satisfy[p.F])) {
ok = false;
}
}
if(ok) {
// all other unknown rows can be alternating in 2 different ways
ans += mod_pow(2, n-total.size());
ans %= MOD;
}
// cols (group by x check all y)
total.clear(), satisfy.clear();
ok = true;
for(int i = 0; i < k; i++) {
total[A[i].x]++;
satisfy[A[i].x] += (A[i].y % 2 == A[i].sgn);
}
for(auto p : total) {
if(!(p.S == satisfy[p.F] || !satisfy[p.F])) {
ok = false;
}
}
if(ok) {
// all other unknown cols can be alternating in 2 different ways
ans += mod_pow(2, m-total.size());
ans %= MOD;
}
// subtract double counted variant when both cols and rows are alternating
// if it's possible to obtain it (only if k == 0 both are possible)
// (checker pattern)
if(k == 0) {
ans = (ans-2+MOD)%MOD;
}
else {
int cnt = 0;
for(int i = 0; i < k; i++) {
cnt += (A[i].x % 2 == A[i].y % 2) == A[i].sgn;
}
if(cnt == 0 || cnt == k) {
ans = (ans-1+MOD)%MOD;
}
}
cout << ans << "\n";
return 0;
}
/*
+ .
. .
2 2 1
+ 1 1
. .
. .
2 2 0
*/
