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;
long long n, k, m, l[500005], r[500005], first[500005], p[500005], v[500005],
    dp[500005], predp[500005];
void solve() {
  long long ans = 1;
  cin >> n >> k >> m;
  for (long long i = 0; i < m; i++) cin >> l[i] >> r[i] >> first[i];
  for (long long bit = 0; bit < k; bit++) {
    for (long long i = 1; i < n + 1; i++) {
      p[i] = 0;
      v[i] = 0;
    }
    for (long long i = 0; i < m; i++) {
      if ((first[i] >> bit) & 1)
        p[l[i]]++, p[r[i] + 1]--;
      else
        v[r[i]] = max(v[r[i]], l[i]);
    }
    for (long long i = 1; i < n + 1; i++) p[i] += p[i - 1];
    for (long long i = 1; i < n + 1; i++) p[i] = p[i] > 0;
    for (long long i = 1; i < n + 1; i++) dp[i] = predp[i] = 0;
    dp[0] = 1;
    predp[0] = (p[1] == 0);
    long long cur = 0;
    for (long long i = 1; i < n + 1; i++) {
      if (v[i] <= cur)
        dp[i] = dp[i - 1] * (1 + !p[i]) % 998244353;
      else {
        cur = v[i];
        dp[i] = (predp[i - 1] - (cur >= 2 ? predp[cur - 2] : 0) + 998244353) %
                998244353;
      }
      predp[i] = (predp[i - 1] + (p[i + 1] == 0 ? dp[i] : 0)) % 998244353;
    }
    ans = (ans * dp[n]) % 998244353;
  }
  cout << ans << '\n';
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
