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;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using ld = long double;
const int mod = 998244353;
const ll inf = 2e9;
const ll INF = 2e18;
const ld EPS = 1e-7;
const int N = 5e5 + 55;
ll l[N], r[N], x[N], n, k, m;
ll d[N], mx[N], dp[N];
void solve() {
  cin >> n >> k >> m;
  for (int i = 1; i <= m; ++i) {
    cin >> l[i] >> r[i] >> x[i];
  }
  ll ans = 1;
  for (int bit = 0; bit < k; ++bit) {
    memset(d, 0, sizeof d);
    memset(mx, 0, sizeof mx);
    for (int i = 1; i <= m; ++i) {
      if ((1 << bit) & x[i]) {
        d[l[i]]++;
        d[r[i] + 1]--;
      } else {
        mx[r[i]] = max(mx[r[i]], l[i]);
      }
    }
    memset(dp, 0, sizeof dp);
    dp[0] = 1;
    ll sum = dp[0], l = 0;
    for (int i = 1; i <= n; ++i) {
      d[i] += d[i - 1];
      mx[i] = max(mx[i], mx[i - 1]);
      if (d[i]) {
        dp[i] = 0;
      } else {
        dp[i] = sum;
      }
      while (l < mx[i]) {
        sum = (sum + mod - dp[l]) % mod;
        ++l;
      }
      sum += dp[i];
      sum %= mod;
    }
    ans *= sum;
    ans %= mod;
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  ;
  int tt = 1;
  while (tt--) {
    solve();
    cout << "\n";
  }
}
