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;
const long long mod = 998244353;
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 3) + (x << 1) + c - '0';
    c = getchar();
  }
  return x * f;
}
int n, k, m, l[500050], r[500050], x[500050], s[500050], lim[500050];
long long ans, dp[500050];
int main() {
  n = read() + 1, k = read(), m = read();
  ans = 1;
  for (int i = 1; i <= m; ++i) {
    l[i] = read(), r[i] = read() + 1, x[i] = read();
  }
  for (int i = 0; i < k; ++i) {
    memset(dp, 0, sizeof(dp));
    memset(s, 0, sizeof(s));
    memset(lim, 0, sizeof(lim));
    for (int j = 1; j <= m; ++j) {
      if ((x[j] >> i) & 1) {
        s[l[j]]++, s[r[j]]--;
      } else {
        lim[r[j]] = max(lim[r[j]], l[j]);
      }
    }
    for (int j = 1; j <= n; ++j) {
      s[j] += s[j - 1];
      lim[j] = max(lim[j], lim[j - 1]);
    }
    dp[0] = 1;
    long long sum = 1;
    int _0 = 0;
    for (int j = 1; j <= n; ++j) {
      while (_0 < lim[j]) sum = (sum - dp[_0] + mod) % mod, ++_0;
      if (s[j]) continue;
      dp[j] = sum;
      sum = (sum + dp[j]) % mod;
    }
    ans = ans * dp[n] % mod;
  }
  cout << ans << endl;
  return 0;
}
