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;
inline long long read() {
  char c;
  long long x = 0;
  long long b = 1;
  do {
    c = getchar();
    if (c == 45) b = -1;
  } while (c > 57 || c < 48);
  do x = x * 10 + c - 48, c = getchar();
  while (c <= 57 && c >= 48);
  x *= b;
  return x;
}
long long n, k, m;
const long long maxn = 5e5 + 5;
const long long mod = 998244353;
long long res = 1, l[maxn], r[maxn], x[maxn], dp[maxn], lst[maxn], dif[maxn];
signed main() {
  n = read(), k = read(), m = read();
  for (long long i = 1; i <= m; i++)
    l[i] = read(), r[i] = read() + 1, x[i] = read();
  for (long long p = 0; p < k; p++) {
    for (long long i = 0; i <= n + 1; i++) lst[i] = dif[i] = dp[i] = 0;
    dp[0] = 1;
    for (long long i = 1; i <= m; i++)
      if (x[i] & (1 << p))
        dif[l[i]]++, dif[r[i]]--;
      else
        lst[r[i]] = max(lst[r[i]], l[i]);
    for (long long i = 2; i <= n + 1; i++)
      dif[i] += dif[i - 1], lst[i] = max(lst[i - 1], lst[i]);
    long long sum = 1, L = 0;
    for (long long i = 1; i <= n + 1; i++) {
      while (L < lst[i]) sum = (mod + sum - dp[L]) % mod, dp[L++] = 0;
      dp[i] = dif[i] ? 0 : sum;
      sum = (sum + dp[i]) % mod;
    }
    res = res * dp[n + 1] % mod;
  }
  cout << res << "\n";
  return 0;
}
