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;
const int mod = 998244353;
int n, k, m, l[N], r[N], x[N];
int s[N], asd[N];
long long dp[N];
long long solve() {
  dp[0] = 1;
  int cur = 0, j = 0;
  long long sum = 1;
  for (int i = 1; i <= n; i++) {
    cur = cur + s[i];
    if (cur == 0) dp[i] = sum;
    while (j < asd[i]) {
      sum = (sum - dp[j] + mod) % mod;
      dp[j] = 0;
      j++;
    }
    sum = (sum + dp[i]) % mod;
  }
  return sum;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> k >> m;
  for (int i = 1; i <= m; i++) cin >> l[i] >> r[i] >> x[i];
  long long res = 1;
  for (int bit = 0; bit < k; bit++) {
    for (int i = 1; i <= n; i++) {
      dp[i] = 0;
      s[i] = 0;
      asd[i] = 0;
    }
    for (int i = 1; i <= m; i++) {
      if (x[i] >> bit & 1) {
        s[l[i]]++;
        s[r[i] + 1]--;
      } else
        asd[r[i]] = max(asd[r[i]], l[i]);
    }
    res = (res * solve()) % mod;
  }
  cout << res;
}
