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;
int n, K, m, dp[500005], mp[500005], d[500005];
vector<pair<int, int> > v[500005];
int calc(int x) {
  for (int i = 0; i <= n; i++) d[i] = 0, dp[i] = 0, mp[i] = 0;
  for (int i = 1; i <= n; i++) {
    for (auto j : v[i]) {
      int l = j.first, p = j.second;
      if (p & (1 << x)) d[l]++, d[i + 1]--;
    }
  }
  for (int i = 2; i <= n; i++) d[i] += d[i - 1];
  int tot = 0, cur = 0, res = 0, al = 1;
  dp[0] = 1;
  for (int i = 1; i <= n; i++) {
    if (d[i])
      mp[i] = tot;
    else
      mp[i] = ++tot;
    if (!d[i]) {
      dp[tot] = (al < res ? al - res + mod : al - res);
      al = (al + dp[tot] >= mod ? al + dp[tot] - mod : al + dp[tot]);
    }
    for (auto j : v[i]) {
      int l = j.first, p = j.second;
      if (!(p & (1 << x))) {
        if (d[l] && d[i] && mp[l] == mp[i]) return 0;
        int pos = d[l] ? mp[l] + 1 : mp[l];
        while (cur < pos)
          res = (res + dp[cur] >= mod ? res + dp[cur] - mod : res + dp[cur]),
          cur++;
      }
    }
  }
  int ans = 0;
  for (int i = cur; i <= tot; i++)
    ans = (ans + dp[i] >= mod ? ans + dp[i] - mod : ans + dp[i]);
  return ans;
}
int main() {
  scanf("%d%d%d", &n, &K, &m);
  for (int i = 1; i <= m; i++) {
    int l, r, x;
    scanf("%d%d%d", &l, &r, &x);
    v[r].push_back(make_pair(l, x));
  }
  int ans = 1;
  for (int i = 0; i < K; i++) {
    ans = 1ll * ans * calc(i) % mod;
  }
  printf("%d\n", ans);
  return 0;
}
