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 maxn = 5e5 + 5;
const int INF = 0x3f3f3f3f;
const long long mod = 998244353;
int n, k, m;
int x[maxn], l[maxn], r[maxn];
long long dp[maxn];
int a[maxn], b[maxn];
int main() {
  scanf("%d%d%d", &n, &k, &m);
  for (int i = 1; i <= m; ++i) scanf("%d%d%d", &l[i], &r[i], &x[i]);
  long long ans = 1;
  dp[0] = 1;
  for (int t = 0; t < k; t++) {
    for (int i = 1; i <= n + 1; i++) a[i] = 0, b[i] = 0;
    for (int i = 1; i <= m; i++) {
      if (x[i] >> t & 1)
        a[l[i]]++, a[r[i] + 1]--;
      else
        b[r[i] + 1] = max(b[r[i] + 1], l[i]);
    }
    for (int i = 1; i <= n + 1; i++) {
      a[i] += a[i - 1];
      b[i] = max(b[i], b[i - 1]);
    }
    for (int i = 1; i <= n + 1; i++) {
      if (!a[i]) {
        dp[i] = dp[i - 1];
        if (b[i]) dp[i] = (dp[i] - dp[b[i] - 1] + mod) % mod;
      } else
        dp[i] = 0;
      dp[i] = (dp[i - 1] + dp[i]) % mod;
    }
    ans = ans * (dp[n + 1] - dp[n] + mod) % mod;
  }
  cout << ans << endl;
}
