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 long long mod = 998244353;
int dp[5 << 17];
vector<pair<pair<int, int>, int> > A;
int main() {
  scanf("%d%d%d", &N, &K, &M);
  for (int i = 0; i < M; i++) {
    int l, r, x;
    scanf("%d%d%d", &l, &r, &x);
    l--;
    A.push_back(make_pair(make_pair(l, r), x));
  }
  long long ans = 1;
  for (int z = 0; z < K; z++) {
    vector<pair<int, int> > y;
    vector<int> x(N + 1);
    for (pair<pair<int, int>, int> p : A) {
      int l = p.first.first, r = p.first.second;
      if (p.second >> z & 1) {
        x[l]++;
        x[r]--;
      } else {
        y.push_back(make_pair(r, l));
      }
    }
    sort(y.begin(), y.end());
    long long cumsum = 1;
    int cnt = 0;
    int iy = 0, il = 0;
    dp[0] = 1;
    for (int i = 0; i < N; i++) {
      cnt += x[i];
      if (cnt > 0)
        dp[i + 1] = 0;
      else {
        dp[i + 1] = cumsum;
        cumsum *= 2;
        if (cumsum >= mod) cumsum -= mod;
      }
      while (iy < y.size() && y[iy].first == i + 1) {
        while (il <= y[iy].second) {
          cumsum += mod - dp[il++];
          if (cumsum >= mod) cumsum -= mod;
        }
        iy++;
      }
    }
    ans = ans * cumsum % mod;
  }
  printf("%lld\n", ans);
}
