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 N = 1e6 + 10;
const long long MOD = 998244353;
int l[N], r[N], x[N], n, m, k, pre[N], clo[N];
long long f[N];
int main() {
  scanf("%d%d%d", &n, &k, &m);
  for (int i = 0; i < m; ++i) scanf("%d%d%d", l + i, r + i, x + i);
  long long ans = 1;
  for (int j = 0; j < k; ++j) {
    for (int i = 0; i <= n; ++i) pre[i] = 0, clo[i] = 0, f[i] = 0;
    for (int i = 0; i < m; ++i)
      if (x[i] & 1 << j)
        ++pre[l[i]], --pre[r[i] + 1];
      else
        clo[r[i]] = max(clo[r[i]], l[i]);
    for (int i = 1; i <= n + 1; ++i)
      pre[i] += pre[i - 1], clo[i] = max(clo[i], clo[i - 1]);
    for (int i = 1; i <= n + 1; ++i) {
      if (pre[i] > 0)
        f[i] = 0;
      else
        f[i] = (f[i - 1] + (clo[i - 1] ? -f[clo[i - 1] - 1] : 1) + MOD) % MOD;
      if (i <= n) f[i] = (f[i] + f[i - 1] + MOD) % MOD;
    }
    ans = ans * f[n + 1] % MOD;
  }
  printf("%lld\n", ans);
  return 0;
}
