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 MAXK = 30;
const int MOD = 998244353;
const int MAXN = 5E5;
tuple<int, int, int> conditions[MAXN];
int dp[MAXN + 1];
int oneEnd[MAXN + 1];
int zeroEnd[MAXN + 1];
int solve(int n, int m, int bit) {
  dp[0] = 1;
  int ones = 0;
  int cur = 0;
  fill(oneEnd, oneEnd + n + 1, 0);
  fill(zeroEnd, zeroEnd + n + 1, 0);
  int left = 0;
  int ptr = 0;
  int sum = 1;
  int trueEnd = 0;
  for (int i = 1; i <= n; i++) {
    while (cur < m && get<0>(conditions[cur]) == i) {
      int end = get<1>(conditions[cur]);
      if (get<2>(conditions[cur]) & (1 << bit)) {
        ones++;
        oneEnd[end]++;
      } else {
        trueEnd = max(trueEnd, get<0>(conditions[cur]));
        zeroEnd[end] = max(zeroEnd[end], get<0>(conditions[cur]));
      }
      cur++;
    }
    ones -= oneEnd[i - 1];
    left = max(left, zeroEnd[i - 1]);
    while (ptr < left) {
      sum = sum - dp[ptr] + MOD;
      if (sum >= MOD) sum -= MOD;
      ptr++;
    }
    if (ones)
      dp[i] = 0;
    else {
      dp[i] = sum;
      sum += dp[i];
      if (sum >= MOD) sum -= MOD;
    }
  }
  int ans = 0;
  for (int i = n; i >= trueEnd; i--) {
    ans += dp[i];
    if (ans >= MOD) ans -= MOD;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, k, m;
  cin >> n >> k >> m;
  for (int i = 0; i < m; i++) {
    int l, r, x;
    cin >> l >> r >> x;
    conditions[i] = make_tuple(l, r, x);
  }
  sort(conditions, conditions + m);
  int ans = 1;
  for (int i = 0; i < k; i++) {
    ans = (int)(((long long)ans * solve(n, m, i)) % MOD);
  }
  cout << ans << "\n";
  return 0;
}
