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 mod = 998244353, ans = 1, sum = 0;
int n, k, m, db[500002];
long long un[500001];
vector<pair<pair<int, int>, int> > sz;
vector<pair<int, int> > kell;
long long solve(int x) {
  int pr = 0, ue = 0;
  kell.clear();
  for (int i = 0; i <= n; i++) un[i] = 0, db[i] = 0;
  for (int i = 0; i < m; i++) {
    int fi = sz[i].first.second, se = sz[i].first.first, id = sz[i].second;
    if (id & x)
      db[fi]++, db[se + 1]--;
    else
      kell.push_back({se, fi});
  }
  kell.push_back({n + 1, n + 1});
  sort(kell.begin(), kell.end());
  for (int i = 1; i <= n; i++) db[i] += db[i - 1];
  un[0] = 1, sum = 1;
  for (int i = 1; i <= n; i++) {
    int maxi = 0;
    while (kell[pr].first == i) maxi = max(maxi, kell[pr].second), pr++;
    if (db[i] == 0) un[i] = sum, sum *= 2, sum %= mod;
    for (int i = ue; i < maxi; i++) sum -= un[i], un[i] = 0;
    ue = max(ue, maxi), sum += (mod * mod), sum %= mod;
  }
  return sum;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> k >> m;
  for (int i = 1; i <= m; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    sz.push_back({{b, a}, c});
  }
  sz.push_back({{n + 1, n + 1}, 0});
  sort(sz.begin(), sz.end());
  for (int j = 1; j < (1 << k); j *= 2) {
    ans *= solve(j), ans %= mod;
  }
  cout << ans << endl;
  return 0;
}
