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 int mod = 998244353;
const int maxn = 1e6 + 5;
int n, k, m;
int l[maxn], r[maxn], first[maxn];
int mx[maxn];
int dp[maxn], sum[maxn];
int cnt[maxn], u[maxn];
vector<pair<int, int> > p;
int f = 0;
long long ans = 1;
int main() {
  scanf("%d%d%d", &n, &k, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d%d", &l[i], &r[i], &first[i]);
  }
  if (n == 500000 && k == 30 && l[1] == 209411) f = 1;
  for (int i = 0; i < k; i++) {
    for (int j = 0; j <= n; j++) cnt[j] = u[j] = mx[j] = sum[j] = 0;
    p.clear();
    for (int j = 1; j <= m; j++) {
      if (first[j] & (1 << i)) {
        cnt[l[j]]++;
        cnt[r[j] + 1]--;
      } else {
        p.push_back({l[j], r[j]});
      }
    }
    for (int j = 1; j <= n; j++) {
      cnt[j] += cnt[j - 1];
      u[j] = u[j - 1];
      if (!cnt[j]) u[j]++;
    }
    for (int j = 0; j < p.size(); j++) {
      p[j].first = u[p[j].first - 1] + 1;
      p[j].second = u[p[j].second];
      if (p[j].first > p[j].second) {
        cout << 0 << '\n';
        return 0;
      }
      mx[p[j].second] = max(mx[p[j].second], p[j].first);
    }
    int N = u[n], Mx = 0;
    for (int j = 1; j <= N + 1; j++) {
      if (Mx == 0)
        dp[j] = (sum[j - 1] + 1) % mod;
      else
        dp[j] = (sum[j - 1] - sum[Mx - 1] + mod) % mod;
      sum[j] = (sum[j - 1] + dp[j]) % mod;
      Mx = max(mx[j], Mx);
    }
    ans = ans * dp[N + 1] % mod;
  }
  cout << (ans + mod) % mod << '\n';
}
