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>
#pragma GCC optimize(3, "Ofast", "inline")
using namespace std;
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
const long long maxn = (long long)1e6 + 5;
const long long mod = 998244353;
const long long inf = 0x3f3f3f3f;
long long T = 1;
long long a[maxn], l[maxn], r[maxn], x[maxn], pos[maxn], dp[maxn];
void solve() {
  long long n, k, m;
  cin >> n >> k >> m;
  for (long long i = 1; i <= m; ++i) cin >> l[i] >> r[i] >> x[i], ++r[i];
  long long ans = 1;
  for (long long p = 0; p <= k - 1; ++p) {
    for (long long i = 1; i <= n + 1; ++i) pos[i] = a[i] = 0;
    for (long long i = 1; i <= m; ++i) {
      if (x[i] & (1 << p))
        ++a[l[i]], --a[r[i]];
      else
        pos[r[i]] = max(pos[r[i]], l[i]);
    }
    for (long long i = 2; i <= n + 1; ++i)
      a[i] += a[i - 1], pos[i] = max(pos[i], pos[i - 1]);
    dp[0] = 1;
    long long sum = 1, l = 0;
    for (long long i = 1; i <= n + 1; ++i) {
      while (l < pos[i]) sum = (sum - dp[l] + mod) % mod, dp[l] = 0, ++l;
      dp[i] = a[i] ? 0 : sum;
      sum = (sum + dp[i]) % mod;
    }
    ans = ans * dp[n + 1] % mod;
  }
  cout << ans;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  while (T--) solve();
  return 0;
}
