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;
const int N = 5e5 + 5;
vector<int> end_points[N];
int pre[N], dp[N], predp[N];
int L[N], R[N], X[N];
int main() {
  int n, k, m;
  cin >> n >> k >> m;
  for (int i = 1; i <= m; i++) {
    cin >> L[i] >> R[i] >> X[i];
    end_points[R[i]].push_back(i);
  }
  int ans = 1;
  for (int b = 0; b < k; b++) {
    for (int i = 1; i <= n; i++) {
      pre[i] = 0;
    }
    for (int i = 1; i <= m; i++) {
      if ((X[i] >> b) & 1) {
        pre[L[i]]++;
        pre[R[i] + 1]--;
      }
    }
    for (int i = 1; i <= n + 1; i++) {
      pre[i] += pre[i - 1];
    }
    dp[0] = 1;
    int lmax = 0;
    predp[0] = 1;
    for (int i = 1; i <= n + 1; i++) {
      for (int idx : end_points[i - 1]) {
        if (!((X[idx] >> b) & 1)) lmax = max(lmax, L[idx]);
      }
      if (pre[i]) {
        dp[i] = 0;
        predp[i] = predp[i - 1];
      } else {
        dp[i] = ((predp[i - 1] - predp[lmax] + mod) % mod + dp[lmax]) % mod;
        predp[i] = (predp[i - 1] + dp[i]) % mod;
      }
    }
    ans = (ans * 1LL * dp[n + 1]) % mod;
  }
  cout << ans;
  return 0;
}
