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 = 5e5 + 5, K = 30, mod = 998244353;
int n, k, m;
pair<pair<int, int>, int> s[N];
int p[N], one[N];
int dp[N], sum[N];
int solve(int t) {
  memset(p, 0, sizeof p);
  memset(one, 0, sizeof one);
  for (int i = 1; i <= m; i++) {
    int l = s[i].first.first, r = s[i].first.second, x = s[i].second;
    if (x >> t & 1)
      one[l]++, one[r + 1]--;
    else
      p[r] = max(p[r], l);
  }
  for (int i = 1, j = 0; i <= n + 1; i++) {
    one[i] += one[i - 1];
    dp[i] = 0;
    if (!one[i])
      if (!j)
        dp[i] = (sum[i - 1] + 1) % mod;
      else
        dp[i] = ((sum[i - 1] - sum[j - 1]) % mod + mod) % mod;
    sum[i] = (sum[i - 1] + dp[i]) % mod;
    j = max(j, p[i]);
  }
  return dp[n + 1];
}
int main() {
  scanf("%d %d %d", &n, &k, &m);
  for (int i = 1; i <= m; i++) {
    int l, r, x;
    scanf("%d %d %d", &l, &r, &x);
    s[i] = {{l, r}, x};
  }
  int res = 1;
  for (int i = 0; i < k; i++) res = 1ll * res * solve(i) % mod;
  printf("%d\n", res);
  return 0;
}
