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;
int n, k, m;
const int N = (int)5e5 + 1;
struct Cond {
  int l;
  int r;
  int x;
};
Cond a[N];
const int MOD = 998244353;
long long mod(long long x) {
  x %= MOD;
  if (x < 0) x += MOD;
  return x;
}
int val[N];
int ls[N];
int dp[N];
long long solve(int k) {
  memset(ls, -1, sizeof ls);
  memset(val, 0, sizeof val);
  for (int j = 0; j < m; j++) {
    if ((a[j].x >> k) & 1) {
      val[a[j].l]++;
      val[a[j].r + 1]--;
    } else {
      ls[a[j].r] = max(ls[a[j].r], a[j].l);
    }
  }
  int j = -1;
  for (int i = 0; i < n; i++) {
    if (!val[i]) {
      dp[i + 1] = mod(dp[i] + dp[i] + (j == -1 ? 1 : -dp[j]));
    } else {
      dp[i + 1] = dp[i];
    }
    val[i + 1] += val[i];
    j = max(j, ls[i]);
  }
  return mod(dp[n] + (j == -1 ? 1 : -dp[j]));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> k >> m;
  for (int i = 0; i < m; i++) {
    cin >> a[i].l >> a[i].r >> a[i].x;
    a[i].l--;
    a[i].r--;
  }
  long long ans = 1;
  for (int i = 0; i < k; i++) {
    ans = mod(ans * solve(i));
    ;
  }
  cout << ans << "\n";
  return 0;
}
