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 maxn = 500007;
int n, m, k, l[maxn], r[maxn], x[maxn];
int add(int u, int v) { return ((u + v) % mod + mod) % mod; }
int solve(int u) {
  vector<int> f, g, dp;
  f.resize(n + 7, 0), g.resize(n + 7, 0), dp.resize(n + 7, 0);
  for (int i = 1; i <= m; ++i) {
    if (x[i] >> u & 1) {
      f[l[i]]++, f[r[i] + 1]--;
    } else {
      g[r[i]] = max(g[r[i]], l[i]);
    }
  }
  dp[0] = 1;
  int p = 0;
  for (int i = 1; i <= n; ++i) {
    if (f[i] > 0) {
      dp[i] = dp[i - 1];
    } else {
      dp[i] = add(dp[i - 1], p ? dp[i - 1] - dp[p - 1] : dp[i - 1]);
    }
    f[i + 1] += f[i];
    p = max(p, g[i]);
  }
  return add(dp[n], p ? -dp[p - 1] : 0);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cin >> n >> k >> m;
  for (int i = 1; i <= m; ++i) {
    cin >> l[i] >> r[i] >> x[i];
  }
  int ret = 1;
  for (int i = 0; i < k; ++i) {
    ret = 1ll * ret * solve(i) % mod;
  }
  cout << ret;
  return 0;
}
