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 <bits/stdc++.h>
#define mset(a, b) memset(a, b, sizeof(a))
#define mcpy(a, b) memcpy(a, b, sizeof(a))
#define rg register
using namespace std;
typedef long long LL;
const int N = 22;
const int MAXS = (1 << 20) + 5;
const int MOD = 1e9 + 7;

template <typename T> inline void read(T &AKNOI) {
    T x = 0, flag = 1;
    char ch = getchar();
    while (!isdigit(ch)) {
        if (ch == '-') flag = -1;
        ch = getchar();
    }
    while (isdigit(ch)) {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    AKNOI = flag * x;
}

int n, m, K, S, must[N][N];
int nxt[MAXS][N], dp[2][N][MAXS], ans;

inline void Inc(int &x, int y) {
	x += y; if (x >= MOD) x -= MOD;
}

void init() {
	read(n); read(m); read(K); 
	--n; S = (1 << n);
	mset(must, -1);
	for (int i = 1, a, b; i <= K; ++i) {
		read(a); read(b);
		read(must[a][b - 1]);
	}
	for (int s = 0; s < S; ++s) {
		int lst = -1;
		for (int i = n - 1; ~i; --i) {
			if ((s >> i) & 1) lst = i;
			nxt[s][i] = lst;
		}
	}
}

void solve() {
	int pre = 0, cur = 1;
	dp[pre][n][0] = 1;
	for (int i = 1; i <= m; ++i) {
		mset(dp[cur], 0);
		for (int s = 0; s < S; ++s) {
			dp[cur][0][s] = dp[pre][n][s];
		}
		for (int j = 0; j < n; ++j) {
			for (int s = 0; s < S; ++s) {
				if (!dp[cur][j][s]) continue;
				for (int b = 0; b < 2; ++b) {
					if (~must[i][j] && must[i][j] != b) continue;
					int k = (s >> j) & 1;
					if (k > b) continue;
					if (k == b) {
						Inc(dp[cur][j + 1][s], dp[cur][j][s]);
					} else {
						int t = nxt[s][j];
						if (t == -1) {
							Inc(dp[cur][j + 1][s ^ (1 << j)], dp[cur][j][s]);
						} else {
							Inc(dp[cur][j + 1][s ^ (1 << j) ^ (1 << t)], dp[cur][j][s]);
						}
					}
				}
			}
		}
		swap(pre, cur);
	}
	for (int s = 0; s < S; ++s) {
		Inc(ans, dp[pre][n][s]);
	}
	printf("%d\n", ans);
}

int main() {
	init();
	solve();
	return 0;
}