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 mod = 998244353;
long long z = 1;
int read() {
  int x, f = 1;
  char ch;
  while (ch = getchar(), ch < '0' || ch > '9')
    if (ch == '-') f = -1;
  x = ch - '0';
  while (ch = getchar(), ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + ch - 48;
  return x * f;
}
int cnt[500005], mx[500005], f[500005], g[500005];
struct node {
  int l, r, x;
} d[500005];
int main() {
  int t, i, j, k, n, m, l, r, ans = 1, p;
  n = read();
  k = read();
  m = read();
  for (i = 1; i <= m; i++) d[i].l = read(), d[i].r = read(), d[i].x = read();
  for (t = 0; t < k; t++) {
    for (i = 1; i <= n + 1; i++) f[i] = g[i] = mx[i] = cnt[i] = 0;
    for (i = 1; i <= m; i++) {
      l = d[i].l;
      r = d[i].r;
      if (1 << t & d[i].x)
        cnt[l]++, cnt[r + 1]--;
      else
        mx[r] = max(mx[r], l);
    }
    for (i = 1; i <= n; i++) cnt[i] += cnt[i - 1];
    f[0] = g[0] = 1;
    p = 0;
    for (i = 1; i <= n; i++) {
      g[i] = g[i - 1];
      if (!cnt[i]) {
        f[i] = g[i - 1];
        g[i] = (g[i] + f[i]) % mod;
      }
      while (p < mx[i]) g[i] = (g[i] - f[p]) % mod, p++;
    }
    ans = z * ans * g[n] % mod;
  }
  ans = (ans % mod + mod) % mod;
  printf("%d", ans);
  return 0;
}
