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 N = 5e5 + 100, mod = 998244353;
int n, m, K, dp[N], s[N], diff[N], ans;
struct block {
  int l, r, a;
} p[N];
vector<int> pot[N];
int main() {
  scanf("%d%d%d", &n, &K, &m);
  for (int i = 1; i <= m; ++i) {
    scanf("%d%d%d", &p[i].l, &p[i].r, &p[i].a);
    pot[p[i].r].push_back(i);
  }
  ans = 1;
  for (int i = 0, lmin; i < K; ++i) {
    lmin = 0;
    for (int j = 0; j <= n + 1; ++j) diff[j] = 0, dp[j] = 0;
    for (int j = 1; j <= m; ++j)
      if (p[j].a & 1) diff[p[j].l] += 1, diff[p[j].r + 1] -= 1;
    s[0] = 1;
    for (int j = 1; j <= n + 1; ++j) {
      diff[j] += diff[j - 1];
      for (auto x : pot[j - 1])
        if (!(p[x].a & 1)) lmin = max(lmin, p[x].l);
      if (diff[j]) {
        s[j] = s[j - 1], dp[j] = 0;
        continue;
      }
      dp[j] = (s[j - 1] + mod - (lmin ? s[lmin - 1] : 0)) % mod,
      s[j] = (s[j - 1] + dp[j]) % mod;
    }
    for (int j = 1; j <= m; ++j) p[j].a >>= 1;
    ans = 1ll * ans * dp[n + 1] % mod;
  }
  printf("%d", ans);
  return 0;
}
