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;
long long n, k, m, dp[500002][2], mx[500002], l[500002], r[500002], sum[500002],
    smen[500002], x[500002];
const long long mod = 998244353;
long long cnt(long long bit) {
  for (long long i = 1; i <= n; i++) smen[i] = mx[i] = dp[i][0] = 0;
  for (long long i = 1; i <= m; i++) {
    if (x[i] & (1 << bit)) {
      smen[l[i]]++;
      smen[r[i] + 1]--;
    } else
      mx[r[i]] = max(mx[r[i]], l[i]);
  }
  for (long long i = 1; i <= n; i++) mx[i] = max(mx[i - 1], mx[i]);
  for (long long i = 1; i <= n; i++) {
    smen[i] += smen[i - 1];
    if (!smen[i]) {
      if (mx[i - 1] == 0)
        dp[i][0] = (sum[i - 1] + 1) % mod;
      else
        dp[i][0] = (sum[i - 1] - sum[mx[i - 1] - 1] + mod) % mod;
    }
    sum[i] = (sum[i - 1] + dp[i][0]) % mod;
  }
  if (mx[n] == 0)
    return sum[n] + 1;
  else
    return (sum[n] - sum[mx[n] - 1] + mod) % mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k >> m;
  for (long long i = 1; i <= m; i++) {
    cin >> l[i] >> r[i] >> x[i];
  }
  long long ans = 1;
  for (long long i = 0; i < k; i++) ans = 1ll * ans * cnt(i) % mod;
  cout << ans;
  return 0;
}
