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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("inline")
#pragma GCC optimize("-ffast-math")
#pragma GCC optimize("inline-small-functions")
#pragma GCC optimize("-finline-small-functions")
#pragma GCC optimize("-fexpensive-optimizations")
#pragma GCC optimize("inline-functions-called-once")
using namespace std;
int i, i1, j, k, k1, t, n, m, flag[10], a, b, l, r, x, s, sz, d[500010],
    e[500010], f[500010];
long long res, dp[500010];
int mk[500010];
vector<array<int, 2>> adj[2];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k >> m;
  for (i = 0; i < m; i++) {
    cin >> d[i] >> e[i] >> f[i];
  }
  res = 1;
  for (b = 0; b < k; b++) {
    adj[0].clear();
    adj[1].clear();
    for (i = 0; i < m; i++) adj[(f[i] >> b) & 1].push_back({e[i], d[i]});
    sort(adj[0].begin(), adj[0].end());
    for (i = 1; i <= n + 1; i++) mk[i] = 0;
    for (auto u : adj[1]) {
      mk[u[1]]++;
      mk[u[0] + 1]--;
    }
    for (i = 1; i <= n + 1; i++) mk[i] += mk[i - 1];
    s = adj[0].size();
    x = 0;
    l = 0;
    dp[0] = 1;
    for (i = 1; i <= n + 1; i++) {
      if (mk[i]) {
        dp[i] = dp[i - 1];
        continue;
      }
      while (l < s && adj[0][l][0] < i) {
        x = max(x, adj[0][l][1]);
        l++;
      }
      dp[i] = (2 * dp[i - 1] - dp[x - 1] + 998244353) % 998244353;
    }
    res *= (dp[n + 1] - dp[n] + 998244353);
    res %= 998244353;
  }
  cout << res << "\n";
  return 0;
}
