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;
const long long mod = 998244353;
int r[maxn], n, k, m;
int s[maxn], dp[maxn];
struct point {
  int l, r, x;
  point(int l = 0, int r = 0, int x = 0) : l(l), r(r), x(x) {}
} p[maxn];
inline long long dec(long long x, long long y) {
  return x >= y ? x - y : x - y + mod;
}
inline long long add(long long x, long long y) {
  return x + y >= mod ? x + y - mod : x + y;
}
inline long long solve(int t) {
  memset(s, 0, sizeof(s));
  memset(dp, 0, sizeof(dp));
  memset(r, 0, sizeof(r));
  for (int i = 1; i <= m; i++) {
    if ((p[i].x >> t) & 1) {
      s[p[i].l]++;
      s[p[i].r + 1]--;
    } else
      r[p[i].r + 1] = max(r[p[i].r + 1], p[i].l);
  }
  for (int i = 1; i <= n + 1; i++) s[i] += s[i - 1];
  for (int i = 1; i <= n + 1; i++) r[i] = max(r[i], r[i - 1]);
  dp[0] = 1;
  int j = 0, res = 1;
  for (int i = 1; i <= n + 1; i++) {
    if (s[i]) {
      dp[i] = 0;
      continue;
    }
    while (j < r[i]) {
      res = dec(res, dp[j]);
      j++;
    }
    dp[i] = res;
    res = add(res, dp[i]);
  }
  return dp[n + 1];
}
int main() {
  cin >> n >> k >> m;
  for (int i = 1; i <= m; i++) scanf("%d%d%d", &p[i].l, &p[i].r, &p[i].x);
  long long ans = 1;
  for (int i = 0; i < k; i++) ans = solve(i) * ans % mod;
  cout << ans << endl;
  return 0;
}
