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 + 10;
const int MOD = 998244353;
long long mul(long long x, long long y) { return x * y % MOD; }
long long add(long long x, long long y) {
  return (x + y >= MOD ? x + y - MOD : x + y);
}
long long sub(long long x, long long y) {
  return (x < y ? x + MOD - y : x - y);
}
int n, K, m;
int l[N], r[N], x[N];
int a[N], b[N];
long long f[N];
long long ans = 1;
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> K >> m;
  for (int i = 1; i <= m; i++) cin >> l[i] >> r[i] >> x[i];
  for (int k = 0; k < K; k++) {
    memset(a, 0, sizeof a);
    memset(b, 0, sizeof b);
    memset(f, 0, sizeof f);
    for (int i = 1; i <= m; i++) {
      if ((1 << k) & x[i])
        a[l[i]]++, a[r[i] + 1]--;
      else
        b[r[i] + 1] = max(b[r[i] + 1], l[i]);
    }
    for (int i = 2; i <= n + 1; i++)
      a[i] += a[i - 1], b[i] = max(b[i], b[i - 1]);
    f[0] = 1;
    long long sum = 1, l = 0;
    for (int i = 1; i <= n + 1; i++) {
      while (l < b[i]) sum = sub(sum, f[l]), l++;
      f[i] = (a[i] ? 0 : sum), sum = add(sum, f[i]);
    }
    ans = mul(ans, f[n + 1]);
  }
  cout << ans << endl;
  return 0;
}
