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 P = 998244353;
const int N = 5e5 + 10;
int a[N], L[N], dp[N];
int qpow(int a, long long b) {
  int res = 1;
  for (; b; b >>= 1) {
    if (b & 1) {
      res = 1ll * res * a % P;
    }
    a = 1ll * a * a % P;
  }
  return res;
}
int main() {
  int n, k, m;
  cin >> n >> k >> m;
  vector<tuple<int, int, int>> seg(m);
  for (auto& i : seg) {
    cin >> get<0>(i) >> get<1>(i) >> get<2>(i);
  }
  int res = 1;
  for (int b = k - 1; b >= 0; b--) {
    memset(a, 0, sizeof(a));
    memset(L, 0, sizeof(L));
    memset(dp, 0, sizeof(dp));
    for (auto& i : seg) {
      int l, r, x;
      tie(l, r, x) = i;
      if (x & (1 << b))
        a[l]++, a[r + 1]--;
      else
        L[r] = max(L[r], l);
    }
    int sum = 1, p = 0;
    dp[0] = 1;
    for (int i = 1; i <= n; i++) {
      a[i] += a[i - 1];
      if (!a[i]) dp[i] = sum, sum = sum * 2 % P;
      while (p < L[i]) sum = (sum - dp[p++]) % P;
    }
    res = 1ll * res * sum % P;
  }
  res = (res + P) % P;
  cout << res;
}
