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 f[500010], d[500010], ans = 1, n, m, mp[500010], k;
vector<pair<int, int> > lim[500010];
int calc(int x) {
  for (int i = 1; i <= n; i++) f[i] = d[i] = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < (int)lim[i].size(); j++)
      if ((lim[i][j].second >> x) & 1) d[lim[i][j].first]++, d[i + 1]--;
  for (int i = 1; i <= n; i++) d[i] += d[i - 1];
  int sum = 1, sum2 = 0, tot = 0, cur = 0;
  f[0] = 1;
  for (int i = 1; i <= n; i++) {
    if (!d[i])
      mp[i] = ++tot;
    else
      mp[i] = tot;
    if (!d[i])
      f[tot] = (sum - sum2 + 998244353) % 998244353,
      (sum += f[tot]) %= 998244353;
    for (int j = 0; j < (int)lim[i].size(); j++)
      if (!((lim[i][j].second >> x) & 1)) {
        if (d[i] && d[lim[i][j].first] && mp[i] == mp[lim[i][j].first])
          return 0;
        int tmp =
            d[lim[i][j].first] ? mp[lim[i][j].first] + 1 : mp[lim[i][j].first];
        while (cur < tmp) (sum2 += f[cur]) %= 998244353, cur++;
      }
  }
  int ans = 0;
  for (int i = cur; i <= tot; i++) (ans += f[i]) %= 998244353;
  return ans;
}
int main() {
  scanf("%d%d%d", &n, &k, &m);
  for (int i = 1, x, y, z; i <= m; i++)
    scanf("%d%d%d", &x, &y, &z), lim[y].push_back(make_pair(x, z));
  for (int i = 0; i < k; i++) ans = 1ll * ans * calc(i) % 998244353;
  printf("%d\n", ans);
}
