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 cansel_sync = [] {
  ios::sync_with_stdio(0);
  cin.tie(0);
  return 0;
}();
const int N = 500010;
const int mod = (0 ? 1000000007 : 998244353);
long long mul(long long a, long long b, long long m = mod) { return a * b % m; }
long long qpow(long long a, long long b, long long m = mod) {
  long long ans = 1;
  for (; b; a = mul(a, a, m), b >>= 1)
    if (b & 1) ans = mul(ans, a, m);
  return ans;
}
long long getinv(long long v, long long m = mod) { return qpow(v, m - 2, m); }
int l[N], r[N], x[N], last[N], v[N], dp[N];
long long sum[N];
int n, m, k;
int main() {
  cin >> n >> k >> m;
  long long ans = 1;
  for (int i = (0), _ = (m); i < _; i++) cin >> l[i] >> r[i] >> x[i];
  for (int i = (0), _ = (k); i < _; i++) {
    dp[0] = sum[0] = 1;
    for (int j = (1), _ = (n + 2); j < _; j++) last[j] = v[j] = 0;
    for (int j = (0), _ = (m); j < _; j++)
      if ((x[j] >> i) & 1) {
        ++v[l[j]];
        --v[r[j] + 1];
      } else
        last[r[j] + 1] = max(last[r[j] + 1], l[j]);
    for (int j = (1), _ = (n + 2); j < _; j++) {
      v[j] += v[j - 1];
      last[j] = max(last[j], last[j - 1]);
    }
    for (int j = (1), _ = (n + 2); j < _; j++) {
      if (!v[j]) {
        dp[j] = sum[j - 1];
        if (last[j]) dp[j] = (dp[j] - sum[last[j] - 1]) % mod;
      } else
        dp[j] = 0;
      sum[j] = (sum[j - 1] + dp[j]) % mod;
    }
    ans = ans * dp[n + 1] % mod;
  }
  cout << (ans + mod) % mod << endl;
}
