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 int p = 998244353;
int n, k, m;
int l[500086], r[500086], x[500086];
int f[500086], maxl[500086];
int der[500086];
int ans = 1;
int main() {
  scanf("%d%d%d", &n, &k, &m);
  k--;
  for (int i = 1; i <= m; i++) scanf("%d%d%d", &l[i], &r[i], &x[i]);
  for (int s = 1 << k; s; s >>= 1) {
    memset(der, 0, sizeof(der));
    memset(maxl, 0, sizeof(maxl));
    memset(f, 0, sizeof(f));
    for (int i = 1; i <= m; i++) {
      if (x[i] & s) {
        der[l[i]]++, der[r[i] + 1]--;
      } else {
        maxl[r[i] + 1] = max(maxl[r[i] + 1], l[i]);
      }
    }
    for (int i = 1; i <= n + 1; i++) der[i] += der[i - 1];
    for (int i = 1; i <= n + 1; i++) maxl[i] = max(maxl[i], maxl[i - 1]);
    f[0] = 1;
    int j = 0, sum = 1;
    for (int i = 1; i <= n + 1; i++) {
      if (der[i]) continue;
      while (j < maxl[i]) {
        sum += p - f[j++];
        if (sum >= p) sum -= p;
      }
      f[i] = sum;
      sum += f[i];
      if (sum >= p) sum -= p;
    }
    ans = 1ll * ans * f[n + 1] % p;
  }
  printf("%d", ans);
}
