You are given three integers n, k, m and m conditions (l_1, r_1, x_1), (l_2, r_2, x_2), ..., (l_m, r_m, x_m).

Calculate the number of distinct arrays a, consisting of n integers such that: 

  * 0 ≤ a_i < 2^k for each 1 ≤ i ≤ n; 
  * bitwise AND of numbers a[l_i] \& a[l_i + 1] \& ... \& a[r_i] = x_i for each 1 ≤ i ≤ m. 



Two arrays a and b are considered different if there exists such a position i that a_i ≠ b_i. 

The number can be pretty large so print it modulo 998244353.

Input

The first line contains three integers n, k and m (1 ≤ n ≤ 5 ⋅ 10^5, 1 ≤ k ≤ 30, 0 ≤ m ≤ 5 ⋅ 10^5) — the length of the array a, the value such that all numbers in a should be smaller than 2^k and the number of conditions, respectively.

Each of the next m lines contains the description of a condition l_i, r_i and x_i (1 ≤ l_i ≤ r_i ≤ n, 0 ≤ x_i < 2^k) — the borders of the condition segment and the required bitwise AND value on it.

Output

Print a single integer — the number of distinct arrays a that satisfy all the above conditions modulo 998244353.

Examples

Input


4 3 2
1 3 3
3 4 6


Output


3


Input


5 2 3
1 3 2
2 5 0
3 3 3


Output


33

Note

You can recall what is a bitwise AND operation [here](https://en.wikipedia.org/wiki/Bitwise_operation#AND).

In the first example, the answer is the following arrays: [3, 3, 7, 6], [3, 7, 7, 6] and [7, 3, 7, 6].
#include <bits/stdc++.h>
using namespace std;
int add(int x, int y) { return ((x + y) % 998244353 + 998244353) % 998244353; }
int mul(int x, int y) { return (long long)x * y % 998244353; }
int maxL[500005];
int L[500005], R[500005], x[500005];
int dp[500005];
int p[500005];
int solveDp(int n, int m, int k, int bit) {
  for (int i = 1; i <= n; ++i) {
    p[i] = 0;
    maxL[i] = 0;
    dp[i] = 0;
  }
  for (int i = 0; i < m; ++i) {
    if (x[i] & (1 << bit)) {
      p[L[i]] += 1;
      p[R[i] + 1] -= 1;
    } else {
      maxL[R[i]] = max(maxL[R[i]], L[i]);
    }
  }
  for (int i = 1; i <= n; ++i) {
    p[i] += p[i - 1];
  }
  int myLeft = 0;
  dp[0] = 1;
  int sum = dp[0];
  for (int i = 1; i <= n; ++i) {
    if (p[i]) {
    } else {
      dp[i] = sum;
      sum = add(sum, dp[i]);
    }
    while (myLeft < maxL[i]) {
      sum = add(sum, -dp[myLeft]);
      ++myLeft;
    }
  }
  return sum;
}
void solve() {
  int n, k, m;
  scanf("%d %d %d ", &n, &k, &m);
  for (int i = 0; i < m; ++i) {
    scanf("%d %d %d ", &L[i], &R[i], &x[i]);
  }
  int ret = 1;
  for (int i = 0; i < k; ++i) {
    ret = mul(ret, solveDp(n, m, k, i));
  }
  printf("%d\n", ret);
}
int main() {
  solve();
  return 0;
}
