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 = 500010;
int n, k, m;
int l[N], r[N], x[N];
int add[N], dp[N], last[N], cur[N];
bool bad[N];
int solve(int bit) {
  for (int i = 0; i < n; i++) add[i] = 0, last[i] = n, bad[i] = false;
  for (int i = 0; i < m; i++) {
    if (((x[i] >> bit) & 1) == 1)
      add[l[i] - 1]++, add[r[i]]--;
    else
      last[l[i] - 1] = min(last[l[i] - 1], r[i] - 1);
  }
  last[n] = cur[n] = n;
  for (int i = n - 1; i >= 0; i--) {
    cur[i] = min(last[i + 1], cur[i + 1]);
  }
  int sum = 0;
  for (int i = 0; i < n; i++) {
    sum += add[i];
    if (sum > 0) bad[i] = true;
  }
  dp[n] = 1;
  int j = n;
  sum = 1;
  for (int i = n - 1; i >= 0; i--) {
    if (bad[i]) {
      dp[i] = 0;
      continue;
    }
    while (j > cur[i]) {
      sum -= dp[j];
      if (sum < 0) sum += 998244353;
      j--;
    }
    dp[i] = sum;
    sum += dp[i];
    if (sum >= 998244353) sum -= 998244353;
  }
  int tmp = min(cur[0], last[0]), ans = 0;
  for (int i = 0; i <= tmp; i++) {
    ans += dp[i];
    if (ans >= 998244353) ans -= 998244353;
  }
  return ans;
}
int main() {
  scanf("%d%d%d", &n, &k, &m);
  for (int i = 0; i < m; i++) scanf("%d%d%d", &l[i], &r[i], &x[i]);
  int ans = 1;
  for (int i = k - 1; i >= 0; i--) ans = (long long)ans * solve(i) % 998244353;
  cout << ans << endl;
  return 0;
}
