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 T>
inline T read() {
  T x = 0, w = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') w = -w;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  return x * w;
}
template <class T>
inline T read(T& x) {
  return x = read<T>();
}
const int mod = 998244353;
inline int add(int a, int b) { return (a += b) >= mod ? a - mod : a; }
inline int mul(int a, int b) { return (long long)a * b % mod; }
inline int fpow(int a, int b) {
  int ans = 1;
  for (; b; b >>= 1, a = mul(a, a))
    if (b & 1) ans = mul(ans, a);
  return ans;
}
const int N = 5e5 + 10;
struct node {
  int l, r, x;
} a[N];
int V[N], L[N], S[N];
int solve(int n) {
  fill(S + 1, S + n + 1, 0), S[0] = 1;
  int lim = -1;
  for (int i = 1; i <= n; ++i) {
    if (!V[i]) {
      int sum = add(S[i - 1], mod - (lim >= 0 ? S[lim] : 0));
      lim = max(lim, L[i] - 1);
      S[i] = add(S[i - 1], sum);
    } else
      S[i] = S[i - 1];
  }
  return add(S[n], mod - (lim >= 0 ? S[lim] : 0));
}
int fa[N];
int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
int main() {
  int n = read<int>(), K = read<int>(), m = read<int>();
  for (int i = 1; i <= m; ++i) read(a[i].l), read(a[i].r), read(a[i].x);
  int ans = 1;
  for (int c = 0; c < K; ++c) {
    for (int i = 1; i <= n; ++i) fa[i] = i;
    fill(V + 1, V + n + 1, 0);
    for (int i = 1; i <= m; ++i)
      if (a[i].x >> c & 1)
        for (int p = find(a[i].r); p >= a[i].l; p = find(p))
          V[p] = 1, fa[p] = p - 1;
    fill(L + 1, L + n + 1, 0);
    bool flag = 0;
    for (int i = 1; i <= m; ++i)
      if (~a[i].x >> c & 1) {
        int p = find(a[i].r);
        if (!p) {
          flag = 1;
          break;
        }
        L[p] = max(L[p], a[i].l);
      }
    if (flag) {
      ans = 0;
      break;
    }
    ans = mul(ans, solve(n));
  }
  printf("%d\n", ans);
  return 0;
}
