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:
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 21, P = 1e9 + 7;
int n, m, k, ans;
int go[1 << N][N], put[1 << N], dp[1 << N], need1[N * N], need2[N * N];
int main() {
	scanf("%d%d%d", &n, &m, &k);
	while(k--) {
		int a, b, c;
		scanf("%d%d%d", &a, &b, &c);
		--a;
		--b;
		need1[a] |= (1 << b);
		need2[a] |= (1 << b) * c;
	}
	--n;
	memset(go, -1, sizeof(go));
	memset(put, -1, sizeof(put));
	dp[0] = 1;
	for(int i = 0; i < (1 << n); ++i) {
		int id = 0;
		for(int j = 0; j < n; ++j) {
			if(i & (1 << j)) {
				if(j && !(i & (1 << (j - 1)))) {
					go[i][id] = i ^ (1 << j) ^ (1 << (j - 1));	
				}
				++id;
			}
		}
		for(int j =  n - 1; j >= 0; --j) {
			if(!(i & (1 << j))) {
				put[i] = i ^ (1 << j);
				break;
			}
		}
	}
	for(int t = 0; t < m; ++t) {
		for(int i = 0; i < (1 << n) - 1; ++i) {
			if(put[i] != -1) {
				dp[put[i]] = (dp[put[i]] + dp[i]) % P;
//				printf("%d -> %d\n", i, put[i]);
			}
		}
		for(int j = 0; j < n; ++j) {
			for(int i = (1 << n) - 1; i >= 0; --i) {	
				if(go[i][j] != -1) {
//					printf("%d -> %d\n", i, go[i][j]);
					dp[go[i][j]] = (dp[go[i][j]] + dp[i]) % P;
				}
			}
		}
		for(int i = 0; i < (1 << n); ++i) {
			if((i & need1[t]) != need2[t]) {
				dp[i] = 0;
			}
		}
	}
	for(int i = 0; i < (1 << n); ++i) {
		ans = (ans + dp[i]) % P;
	}
	printf("%d\n", ans);
	return 0;
}