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 long long MAXN = 5e5 + 5;
const long long MOD = 998244353;
long long res = 1;
vector<pair<long long, long long>> v1[MAXN];
long long pref[MAXN];
vector<pair<long long, long long>> v2[MAXN];
long long dp[MAXN];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, m, k;
  cin >> n >> k >> m;
  for (long long i = 1; i <= m; i++) {
    long long l, r, x;
    cin >> l >> r >> x;
    v1[l].push_back(make_pair(r, x));
    v2[r + 1].push_back(make_pair(l, x));
  }
  for (long long bit = 0; bit < k; bit++) {
    long long val = (1LL << bit);
    long long cnt = 0;
    long long maxr = 0;
    for (long long i = 1; i <= n + 1; i++) {
      dp[i] = pref[i] = 0;
    }
    dp[0] = 1;
    pref[0] = 1;
    for (long long i = 1; i <= n + 1; i++) {
      pref[i] = pref[i - 1];
      for (auto x : v1[i]) {
        if ((x.second & val)) {
          cnt++;
        }
      }
      for (auto x : v2[i]) {
        if ((x.second & val)) {
          cnt--;
        } else {
          maxr = max(maxr, x.first);
        }
      }
      if (cnt == 0) {
        dp[i] = pref[i - 1] - pref[maxr - 1];
        dp[i] += MOD;
        dp[i] %= MOD;
      }
      pref[i] += dp[i];
      pref[i] %= MOD;
    }
    res = 1LL * res * dp[n + 1];
    res %= MOD;
  }
  cout << res << "\n";
}
