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;

const int Mod = 1e9 + 7, N = 25, MaxMM = (1<<19) + 5;

int n, m, i, j, k, mask, MM, restriction[N], valuare[N], x, y, dir, dp[N][MaxMM];


void add(int &a, int b)
{
    a += b;
    if(a >= Mod) a -= Mod;
}

bool good(int pos, int mask)
{
    return ((mask & restriction[pos]) == valuare[pos]);
}

int main()
{
 //   freopen("input", "r", stdin);
    cin.sync_with_stdio(false);

    cin >> n >> m >> k;
    while(k--)
    {
        cin >> x >> y >> dir;
        restriction[x] |= (1 << (y-1));
        if(dir) valuare[x] |= (1 << (y-1));
    }

    MM = (1<<(n-1)) - 1;
    for(mask = 0; mask <= MM; ++ mask)
        dp[0][mask] = good(1, mask);

    for(i=2; i<=m; ++i)
    {
        for(j=0; j<n-1; ++j)
            for(mask = 0; mask <= MM; ++ mask)
            {
                add(dp[j+1][mask], dp[j][mask]);
                if( !(mask & (1<<j)) )
                {
                    int new_mask = mask & (MM - (1<<(j+1)) + 1);
                    new_mask = new_mask & (-new_mask);
                    add(dp[j+1][ (mask ^ new_mask) | (1<<j) ], dp[j][mask]);
                }
            }

        for(mask = 0; mask <= MM; ++ mask)
            dp[0][mask] = good(i, mask) * dp[n-1][mask];

        for(j=1; j<=n-1; ++j) memset(dp[j], 0, sizeof(dp[j]));
    }

    int ans = 0;
    for(mask = 0; mask <= MM; ++ mask)
            add(ans, dp[0][mask]);

    cout << ans << '\n';
    return 0;
}
