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 N = 5e5 + 7;
const int MOD = 998244353;
vector<pair<int, int>> op[N];
vector<pair<int, int>> cl[N];
int n;
int mxl[N];
void init(int bt) {
  mxl[0] = 0;
  for (int i = 1; i <= n; i++) {
    mxl[i] = mxl[i - 1];
    for (auto [v, l] : cl[i]) {
      if ((v >> bt) & 1) continue;
      mxl[i] = max(mxl[i], l);
    }
  }
}
int dp[N];
int ps[N];
int sm(int l, int r) { return (ps[r] - ps[l] + MOD) % MOD; }
int solve(int bt) {
  init(bt);
  dp[0] = 1;
  ps[0] = 0;
  ps[1] = dp[0];
  int cb = 0;
  for (int i = 1; i <= n; i++) {
    for (auto [v, r] : op[i]) {
      if ((v >> bt) & 1) cb++;
    }
    if (cb) {
      dp[i] = 0;
    } else {
      dp[i] = sm(mxl[i - 1], i);
    }
    ps[i + 1] = (ps[i] + dp[i]) % MOD;
    for (auto [v, l] : cl[i]) {
      if ((v >> bt) & 1) cb--;
    }
  }
  return sm(mxl[n], n + 1);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int k, m;
  cin >> n >> k >> m;
  while (m--) {
    int l, r, x;
    cin >> l >> r >> x;
    op[l].push_back({x, r});
    cl[r].push_back({x, l});
  }
  long long ans = 1;
  for (int i = 0; i < k; i++) {
    int v = solve(i);
    ans *= (long long)v;
    ans %= MOD;
  }
  cout << ans << endl;
}
