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;
template <class C>
void mini(C& a, C b) {
  a = min(a, b);
}
template <class C>
void maxi(C& a, C b) {
  a = max(a, b);
}
const long long mod = 998244353;
int n, k, m;
int l, r, x;
vector<int> b;
vector<long long> dp;
vector<vector<pair<int, int>>> c;
long long g(vector<tuple<int, int, int>>& a, int bit) {
  for (int i = 0; i < n + 2; i++) b[i] = 0;
  for (int i = 0; i < n + 1; i++) dp[i] = 0;
  for (auto t : a) {
    tie(l, r, x) = t;
    if (x & (1 << bit)) b[l]++, b[r]--;
  }
  for (int(i) = 1; i <= n + 1; i++) b[i] += b[i - 1];
  int j = 0;
  long long suma = 1;
  dp[0] = 1;
  int i = 1;
  while (i <= n + 1) {
    for (auto t : c[i]) {
      tie(l, x) = t;
      if (x & (1 << bit)) continue;
      while (j < l) {
        suma -= dp[j];
        suma %= mod;
        j++;
      }
    }
    if (i == n + 1) break;
    if (b[i] == 0) dp[i] = suma;
    suma += dp[i];
    suma %= mod;
    i++;
  }
  return suma;
}
void f() {
  cin >> n >> k >> m;
  b.resize(n + 2);
  dp.resize(n + 1);
  c.resize(n + 2);
  vector<tuple<int, int, int>> a;
  for (int i = 0; i < m; i++) {
    cin >> l >> r >> x;
    r++;
    a.push_back({l, r, x});
    c[r].push_back({l, x});
  }
  long long res = 1;
  for (int i = 0; i < k; i++) res *= g(a, i), res %= mod;
  if (res < 0) res += mod;
  cout << res << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  f();
}
