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>
const int mod = 998244353;
const int N = 1 << 19;
int l[500010], r[500010], x[500010], c[500010], to[500010], s[500010];
int b[N + N], lv[N + N], n, m;
int go(int i) {
  for (int j = 0; j <= n + 2; j++) {
    to[j] = -1;
    c[j] = s[j] = 0;
  }
  for (int j = 0; j < m; j++) {
    if (x[j] >> i & 1) {
      c[l[j]]++;
      c[r[j]]--;
    } else {
      if (to[r[j]] < l[j]) to[r[j]] = l[j];
    }
  }
  int cc = 0, tmp = 0, L = 0;
  s[1] = 1;
  for (int j = 0; j <= n; j++) {
    cc += c[j];
    if (L < to[j] + 1) L = to[j] + 1;
    tmp = s[j + 1] - s[L];
    if (tmp < 0) tmp += mod;
    if (!cc) s[j + 2] = tmp;
    s[j + 2] += s[j + 1];
    if (s[j + 2] >= mod) s[j + 2] -= mod;
  }
  return tmp;
}
int main() {
  int k;
  scanf("%d%d%d", &n, &k, &m);
  for (int i = 0; i < m; i++) {
    scanf("%d%d%d", &l[i], &r[i], &x[i]);
    --l[i];
  }
  int ans = 1;
  for (int i = 0; i < k; i++) {
    ans = (long long)ans * go(i) % mod;
  }
  printf("%d\n", ans);
  return 0;
}
