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>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
using namespace std;
const long long maxn = 5e5 + 100;
const long long mod = 998244353;
long long n, m, k, t;
long long st2[maxn];
long long L[maxn];
long long R[maxn];
long long X[maxn];
vector<long long> open[maxn];
vector<long long> close[maxn];
long long ANS[maxn];
long long solve(long long i) {
  long long q, w, e;
  vector<long long> A(n);
  vector<long long> D(n + 3);
  for (q = 0; q < n; q++) {
    A[q] = ((ANS[q] & st2[i]) != 0);
  }
  long long l = -1, sum = 0;
  D[0] = 0;
  for (q = 0; q < n; q++) {
    if (A[q] != 1) {
      D[q] = sum;
      if (l == -1) {
        D[q]++;
      }
      sum += D[q];
      sum %= mod;
    }
    for (w = 0; w < close[q].size(); w++) {
      if ((X[close[q][w]] & st2[i]) != 0) {
        continue;
      }
      long long ind = close[q][w];
      if (L[ind] > l) {
        while (l < L[ind]) {
          if (l != -1) {
            sum -= D[l];
            sum += mod;
            sum %= mod;
          }
          l++;
        }
      }
    }
  }
  if (l == -1) {
    sum++;
  }
  return sum;
}
int main() {
  long long q, w, e, a, b, c, d;
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  st2[0] = 1;
  for (q = 1; q < 40; q++) {
    st2[q] = st2[q - 1] * 2;
  }
  cin >> n >> k >> m;
  for (q = 0; q < m; q++) {
    cin >> L[q] >> R[q] >> X[q];
    L[q]--;
    R[q]--;
    open[L[q]].push_back(q);
    close[R[q]].push_back(q);
  }
  vector<long long> cnt(k);
  for (q = 0; q < n; q++) {
    for (w = 0; w < open[q].size(); w++) {
      long long first = X[open[q][w]];
      for (e = 0; e < k; e++) {
        if ((st2[e] & first) != 0) {
          cnt[e]++;
        }
      }
    }
    for (w = 0; w < k; w++) {
      if (cnt[w] != 0) {
        ANS[q] += st2[w];
      }
    }
    for (w = 0; w < close[q].size(); w++) {
      long long first = X[close[q][w]];
      for (e = 0; e < k; e++) {
        if ((st2[e] & first) != 0) {
          cnt[e]--;
        }
      }
    }
  }
  long long ans = 1;
  for (q = 0; q < k; q++) {
    ans *= solve(q);
    ans %= mod;
  }
  cout << ans;
  return 0;
}
