Problem: There are N(N+1)/2 dots arranged to form an equilateral triangle whose sides consist of N dots, as shown below. The j-th dot from the left in the i-th row from the top is denoted by (i, j) (1 \leq i \leq N, 1 \leq j \leq i). Also, we will call (i+1, j) immediately lower-left to (i, j), and (i+1, j+1) immediately lower-right to (i, j).

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Takahashi is drawing M polygonal lines L_1, L_2, ..., L_M by connecting these dots. Each L_i starts at (1, 1), and visits the dot that is immediately lower-left or lower-right to the current dots N-1 times. More formally, there exist X_{i,1}, ..., X_{i,N} such that:

* L_i connects the N points (1, X_{i,1}), (2, X_{i,2}), ..., (N, X_{i,N}), in this order.
* For each j=1, 2, ..., N-1, either X_{i,j+1} = X_{i,j} or X_{i,j+1} = X_{i,j}+1 holds.



Takahashi would like to draw these lines so that no part of L_{i+1} is to the left of L_{i}. That is, for each j=1, 2, ..., N, X_{1,j} \leq X_{2,j} \leq ... \leq X_{M,j} must hold.

Additionally, there are K conditions on the shape of the lines that must be followed. The i-th condition is denoted by (A_i, B_i, C_i), which means:

* If C_i=0, L_{A_i} must visit the immediately lower-left dot for the B_i-th move.
* If C_i=1, L_{A_i} must visit the immediately lower-right dot for the B_i-th move.



That is, X_{A_i, {B_i}+1} = X_{A_i, B_i} + C_i must hold.

In how many ways can Takahashi draw M polygonal lines? Find the count modulo 1000000007.

Constraints

* 1 \leq N \leq 20
* 1 \leq M \leq 20
* 0 \leq K \leq (N-1)M
* 1 \leq A_i \leq M
* 1 \leq B_i \leq N-1
* C_i = 0 or 1
* No pair appears more than once as (A_i, B_i).

Input

Input is given from Standard Input in the following format:


N M K
A_1 B_1 C_1
A_2 B_2 C_2
:
A_K B_K C_K


Output

Print the number of ways for Takahashi to draw M polygonal lines, modulo 1000000007.

Examples

Input

3 2 1
1 2 0


Output

6


Input

3 2 2
1 1 1
2 1 0


Output

0


Input

5 4 2
1 3 1
4 2 0


Output

172


Input

20 20 0


Output

881396682

Solution (in CPP):

Solution:
#pragma GCC optimize("O2,unroll-loops")
#pragma GCC target("avx,avx2,popcnt")
#include<bits/stdc++.h>
#define all(x) x.begin(), x.end()
#define pb push_back
using namespace std;
using ll = long long;
using ld = long double;
const int maxn = 1<<19, mod = 1e9 + 7;
int n, m, k, g[23][2], dp[2][maxn][22], go[maxn][22];//[msk][cur_layer][cur_bit]
void pre() {
	for(int i = 0; i < maxn; i++) {
		go[i][19] = -1;
		for(int j = 19; j--;) {
			if(~i >> j & 1) go[i][j] = -1;
			else {
				if(i >> (j+1) & 1) go[i][j] = go[i][j+1];
				else go[i][j] = i;
				if(go[i][j] != -1)
					go[i][j] ^= 3<<j;
			}
                        if(go[i][j] != -1 && go[i][j] >= maxn) go[i][j] = -1;
		}
	}
}
void add(int &a, int b) {
	a += b;
	if(a >= mod) a -= mod;
}
int main() {
	cin.tie(0)->sync_with_stdio(0);
	pre();
	cin >> n >> m >> k;--n;
	for(int a, b, c, i = 0; i < k; i++) {
		cin >> a >> b >> c;--a, --b;
		g[a][c] |= 1<<(n-b-1);
	}
	for(int i = 0; i <= n; i++) {
		dp[0][(1<<i)-1][0] = 1;
	}
	for(int i = 0; i < m; i++) {
		int I = i&1;
		int J = I^1;
		memset(dp[J], 0, sizeof dp[J]);
		for(int j = 0; j < 1<<n; j++) {
			for(int k = 0; k < n; k++) {
				//if(dp[i][j][k]) {
				//	cout << i << " " << j << " " << k << " " << dp[i][j][k] << '\n';
				//}
				add(dp[I][j][k+1], dp[I][j][k]);
				if(go[j][k] != -1) {
					add(dp[I][go[j][k]][k+1], dp[I][j][k]);
				}
			}
			if((j&g[i][1]) != g[i][1] || (j&g[i][0])) continue; 
			add(dp[J][j][0], dp[I][j][n]);
			//cout << i << " - " << j << " -> " << j << "\n";
			for(int k = 0; k < n; k++) if(~j >> k & 1) {
				int t = j|((1<<(k+1))-1);
			//cout << i << " - " << j << " -> " << t << "\n";
				add(dp[J][t][0], dp[I][j][n]);
			}
		}
	}
	cout << dp[m&1][(1<<n)-1][0];
	return 0;
}