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 maxn = 5e5 + 50;
const int mod = 998244353;
int l[maxn], r[maxn], x[maxn], pre[maxn];
int n, k, m;
int dp[maxn], s[maxn], d[maxn];
long long sol(int id) {
  for (int i = (0); i < (n + 1); ++i) pre[i] = 0, dp[i] = s[i] = d[i] = 0;
  for (int i = (0); i < (m); ++i) {
    if (x[i] >> id & 1)
      d[l[i]]++, d[r[i] + 1]--;
    else
      pre[r[i]] = max(pre[r[i]], l[i]);
  }
  dp[0] = s[0] = 1;
  for (int i = (1); i < (n + 1); ++i) {
    pre[i] = max(pre[i], pre[i - 1]), d[i] += d[i - 1];
    if (d[i] > 0)
      dp[i] = 0;
    else if (pre[i - 1] > 0)
      dp[i] = (s[i - 1] - s[pre[i - 1] - 1]) % mod;
    else
      dp[i] = s[i - 1];
    s[i] = (s[i - 1] + dp[i]) % mod;
  }
  return (s[n] - s[pre[n] - 1]) % mod;
}
int main() {
  cin >> n >> k >> m;
  for (int i = (0); i < (m); ++i) scanf("%d%d%d", &l[i], &r[i], &x[i]);
  long long ans = 1;
  for (int i = (0); i < (k); ++i) ans = ans * sol(i) % mod;
  cout << (ans + mod) % mod << endl;
}
