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;
vector<pair<int, int>> vb[30];
vector<pair<int, int>> vw[30];
int bb[505050];
int qq[505050];
long long dd[505050][2];
long long ss[505050];
long long calc(int b) {
  sort(vb[b].begin(), vb[b].end());
  int j = 0, mx = 0;
  for (int i = 1; i <= n; i++) {
    while (j < vb[b].size() && vb[b][j].first == i) {
      mx = max(mx, vb[b][j].second);
      j++;
    }
    bb[i] = i <= mx;
  }
  sort(vw[b].begin(), vw[b].end());
  mx = 0;
  for (int i = 1; i <= n; i++) {
    qq[i] = 0;
  }
  for (int i = 0; i < vw[b].size(); i++) {
    int s = -vw[b][i].second;
    if (s > mx) {
      qq[vw[b][i].first] = s;
      mx = s;
    }
  }
  dd[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    dd[i][0] = 0;
    if (!bb[i]) {
      dd[i][0] += dd[i - 1][0];
      dd[i][0] += dd[i - 1][1];
      dd[i][0] %= 998244353;
    }
    dd[i][1] = 0;
    if (qq[i] == 0) {
      dd[i][1] += dd[i - 1][0];
      dd[i][1] += dd[i - 1][1];
      dd[i][1] %= 998244353;
    } else {
      long long hh =
          ((ss[i - 1] - ss[qq[i] - 1]) % 998244353 + 998244353) % 998244353;
      dd[i][1] += hh;
      dd[i][1] %= 998244353;
    }
    ss[i] = ss[i - 1] + dd[i][0];
    ss[i] %= 998244353;
  }
  return (dd[n][0] + dd[n][1]) % 998244353;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k >> m;
  for (int i = 1; i <= m; i++) {
    int l, r, x;
    cin >> l >> r >> x;
    for (int b = 0; b < k; b++) {
      if (x & (1 << b))
        vb[b].push_back({l, r});
      else
        vw[b].push_back({r, -l});
    }
  }
  long long u = 1;
  for (int b = 0; b < k; b++) {
    u *= calc(b);
    u %= 998244353;
  }
  cout << u << "\n";
}
