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>
using namespace std;

typedef pair<int, int> pii;
typedef long long ll;
typedef vector<int> vi;

#define pb push_back
#define eb emplace_back
#define mp make_pair
#define fi first
#define se second
#define rep(i,n) rep2(i,0,n)
#define rep2(i,m,n) for(int i=m;i<(n);i++)
#define ALL(c) (c).begin(),(c).end()
#define dump(x) cout << #x << " = " << (x) << endl
constexpr ll TEN(int n) { return (n == 0) ? 1 : 10 * TEN(n-1); }

template<class T, class U>
ostream& operator<<(ostream& os, const pair<T, U>& p) {
	os<<"("<<p.first<<","<<p.second<<")";
	return os;
}

template<class T>
ostream& operator<<(ostream& os, const vector<T>& v) {
	os<<"{";
	rep(i, v.size()) {
		if (i) os<<",";
		os<<v[i];
	}
	os<<"}";
	return os;
}

const int MOD = 1000000007;
const int MX = 20;

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

int dir[MX][MX];
int dp[2][1 << (MX-1)];

int main() {
	int N, M, K; cin >> N >> M >> K;
	memset(dir, -1, sizeof(dir));
	rep(i, K) {
		int A, B, C; cin >> A >> B >> C;
		--A; --B;
		dir[A][B] = C;
	}

	int c = 0, f = 1;

	dp[c][0] = 1;

	rep(i, M) {
		rep(j, N-1) {
			rep(mask, (1 << N-1)) {
				dp[f][mask] = 0;
			}
			rep(mask, 1 << (N-1)) {
				if (dp[c][mask] == 0) {
					continue;
				}
				int u = (mask >> j) & 1;

				//same
				if (dir[i][j] != (u^1)) {
					add(dp[f][mask], dp[c][mask]);
				}

				//diff
				if (dir[i][j] != 0 && u == 0) {
					int hbit = mask >> (j+1);
					hbit -= hbit & (-hbit);
					int nmask = (hbit << (j+1)) | (mask & ((1 << j)-1)) | (1 << j);

					add(dp[f][nmask], dp[c][mask]);
				}
			}
			swap(c, f);
		}
	}

	int ans = 0;
	rep(mask, 1 << (N-1)) {
		add(ans, dp[c][mask]);
	}
	cout << ans << endl;

	return 0;
}