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 int n, q;
long long int L[500005], R[500005];
long long int X[500005];
vector<long long int> seq, sq;
long long int solve(long long int w) {
  vector<long long int> dp(n + 2, 0);
  vector<long long int> pref(n + 2, 0);
  vector<bool> must(n + 2, false);
  long long int ans = 1;
  long long int r = -1;
  long long int ptr = 0;
  for (long long int i = 1; i <= n; i++) {
    while (ptr < q && L[seq[ptr]] == i) {
      if (w & X[seq[ptr]]) r = max(r, R[seq[ptr]]);
      ptr++;
    }
    if (r >= i) must[i] = true;
  }
  ptr = 0;
  dp[0] = 0;
  pref[0] = 1;
  long long int l = 0;
  for (long long int i = 1; i <= n + 1; i++) {
    if (!must[i]) {
      if (l != 0)
        dp[i] = (pref[i - 1] - pref[l - 1] + 998244353) % 998244353;
      else
        dp[i] = pref[i - 1];
    } else {
      dp[i] = 0;
    }
    pref[i] = (pref[i - 1] + dp[i]) % 998244353;
    while (ptr < q && R[sq[ptr]] == i) {
      if (!(w & X[sq[ptr]])) {
        l = max(l, L[sq[ptr]]);
      }
      ptr++;
    }
  }
  return dp[n + 1];
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int k, ans = 1;
  cin >> n >> k >> q;
  for (long long int i = 0; i < q; i++)
    cin >> L[i] >> R[i] >> X[i], seq.push_back(i), sq.push_back(i);
  sort((seq).begin(), (seq).end(),
       [&](long long int x, long long int y) { return L[x] < L[y]; });
  sort((sq).begin(), (sq).end(),
       [&](long long int x, long long int y) { return R[x] < R[y]; });
  for (long long int i = 0; i < k; i++) {
    ans = ans * solve(1 << i) % 998244353;
  }
  cout << ans;
}
