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
#include <bits/stdc++.h>
using namespace std;
int n, m, k;
const int N = 23;
const int M = 530000;
const int MOD = 1000000007;
int f[2][M], V[N][N];
int lowbit(int t) {return t & (-t);}
int trans(int t, int p, int c)
{
    if (t & (1 << p))
    {
        if (c == 0) return 1 << (n - 1);
        else return t;
    }
    else
    {
        if (c == 0) return t;
        return (((t >> p) - lowbit(t >> p) + 1) << p) | (t & ((1 << p) - 1));
    }
}
void ADD(int &t, int d)
{
    t += d; if (t >= MOD) t -= MOD;
}
int main()
{
    cin >> n >> m >> k;
    for (int i = 1; i <= m; ++ i)
        for (int j = 0; j < n - 1; ++ j)
            V[i][j] = -1;
    for (int i = 1; i <= k; ++ i)
    {
        int a, b;
        cin >> a >> b; b --;
        cin >> V[a][b];
    }
    int tmp = 0;
    f[0][0] = 1;
    for (int i = 1; i <= m; ++ i)
        for (int j = 0; j < n - 1; ++ j)
        {
            memset(f[!tmp], 0, sizeof f[!tmp]);
            for (int k = 0; k < (1 << (n - 1)); ++ k)
            {
                if (V[i][j] == -1 || V[i][j] == 0) ADD(f[!tmp][trans(k, j, 0)], f[tmp][k]);
                if (V[i][j] == -1 || V[i][j] == 1) ADD(f[!tmp][trans(k, j, 1)], f[tmp][k]);
            }
            tmp ^= 1;
        }
    int ans = 0;
    for (int k = 0; k < (1 << (n - 1)); ++ k) ADD(ans, f[tmp][k]);
    cout << ans;
}
