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 + 10, mod = 998244353;
struct Node {
  int l, r, x;
} a[maxN + 1];
int n, m, k;
int f[maxN + 1];
int s[maxN + 1], L[maxN + 1];
inline int ADD(int x, int y) { return x + y >= mod ? x + y - mod : x + y; }
inline int SUB(int x, int y) { return x - y < 0 ? x - y + mod : x - y; }
inline int solve(int t) {
  for (int i = 0; i <= n + 1; i++) s[i] = f[i] = L[i] = 0;
  for (int i = 1; i <= m; i++)
    if (a[i].x & (1 << t))
      s[a[i].l]++, s[a[i].r + 1]--;
    else
      L[a[i].r] = max(L[a[i].r], a[i].l);
  int sum, now = 0;
  sum = f[0] = 1;
  for (int i = 1; i <= n; i++) {
    s[i] += s[i - 1];
    if (!s[i]) f[i] = sum, sum = ADD(sum, f[i]);
    while (now < L[i]) sum = SUB(sum, f[now]), now++;
  }
  return sum;
}
int main() {
  scanf("%d %d %d", &n, &k, &m);
  for (int i = 1; i <= m; i++) scanf("%d %d %d", &a[i].l, &a[i].r, &a[i].x);
  int ans = 1;
  for (int i = 0; i < k; i++) ans = 1ll * ans * solve(i) % mod;
  printf("%d", ans);
  return 0;
}
