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 MaxN = 5e5 + 5;
const int MaxK = 31;
const int ModM = 998244353;
struct Condition {
  int l;
  int r;
  int x;
  Condition(int l, int r, int x) : l(l), r(r), x(x) {}
};
int t, n, m, k;
vector<Condition> C;
long long mul(long long a, long long b) { return (a * b) % ModM; }
long long add(long long a, long long b) { return (a + b) % ModM; }
long long sub(long long a, long long b) { return (a - b + ModM) % ModM; }
long long Kamui(int x) {
  vector<Condition> Hi;
  vector<Condition> Lo;
  for (auto c : C)
    if (c.x & (1 << x))
      Hi.push_back(c);
    else
      Lo.push_back(c);
  vector<long long> A(n + 2, 0);
  for (auto c : Hi) A[c.l]++, A[c.r + 1]--;
  for (int i = 1; i <= n + 1; ++i) A[i] += A[i - 1];
  vector<int> L(n + 2, 0);
  for (auto c : Lo) L[c.r + 1] = max(L[c.r + 1], c.l);
  for (int i = 1; i <= n + 1; ++i) L[i] = max(L[i - 1], L[i]);
  vector<long long> Dp1(n + 2, 0);
  vector<long long> Dp2(n + 2, 0);
  Dp1[0] = Dp2[0] = 1;
  for (int i = 1; i <= n + 1; ++i) {
    Dp1[i] = Dp2[i - 1];
    if (L[i]) Dp1[i] = sub(Dp1[i], Dp2[L[i] - 1]);
    if (A[i]) Dp1[i] = 0;
    Dp2[i] = add(Dp2[i - 1], Dp1[i]);
  }
  return Dp1[n + 1];
}
int Go() {
  scanf("%d %d %d", &n, &k, &m);
  for (int i = 0; i < m; ++i) {
    int l, r, x;
    scanf("%d %d %d", &l, &r, &x);
    C.push_back(Condition(l, r, x));
  }
  long long Ans = 1;
  for (int i = 0; i < k; ++i) Ans = mul(Ans, Kamui(i));
  cout << Ans << endl;
  return 0;
}
int main() {
  if (0)
    scanf("%d", &t);
  else
    t = 1;
  while (t--) Go();
}
