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 <vector>
using namespace std;

const int N = 1 << 19;
const int MOD = 1e9 + 7;
inline void add(int &a, int b) { a += b; if (a >= MOD) a -= MOD; }

int main() {
        int n, m, k;
        scanf("%d%d%d", &n, &m, &k);
        n --; //attention
        vector<vector<int>> check(m, vector<int> (n, -1));
        for (int i = 0; i < k; i ++) {
                int a, b, c;
                scanf("%d%d%d", &a, &b, &c);
                a --, b --;
                check[a][b] = c;
        }
        vector<vector<int>> dp(2, vector<int> (N, 0));
        dp[0][0] = 1;
        for (int i = 0; i < m; i ++) {
                for (int j = 0; j < n; j ++) {
                        for (int mask = 0; mask < (1 << n); mask ++) {
                                dp[1][mask] = 0;
                        }
                        for (int mask = 0; mask < (1 << n); mask ++) {
                                if (dp[0][mask] == 0) continue;
                                //left
                                if (check[i][j] != 1 && (~mask >> j) & 1) {
                                        add(dp[1][mask], dp[0][mask]);
                                } 
                                //right
                                if (check[i][j] != 0) {
                                        int mask2 = (((mask - (1 << j)) & mask) | (1 << j));
                                        add(dp[1][mask2], dp[0][mask]);
                                }
                        }
                        swap(dp[0], dp[1]);
                }
        }
        int ans = 0;
        for (int mask = 0; mask < (1 << n); mask ++) {
                add(ans, dp[0][mask]);
        }
        printf("%d\n", ans);
        return 0;
}