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 mod = 998244353;
int l[maxn], r[maxn], v[maxn];
int n, k, m;
int pre[maxn];
int mx[maxn];
long long dp[maxn], sum[maxn];
long long add(long long a, long long b) { return ((a + b) % mod + mod) % mod; }
long long work(int idx) {
  memset(pre, 0, sizeof(pre));
  memset(mx, 0, sizeof(mx));
  memset(dp, 0, sizeof(dp));
  for (int i = 1; i <= m; i++) {
    if ((v[i] >> idx) & 1) {
      pre[l[i]] += 1;
      pre[r[i] + 1] -= 1;
    } else {
      mx[r[i]] = max(mx[r[i]], l[i]);
    }
  }
  int real_mx = 0;
  for (int i = 1; i <= n; i++) {
    pre[i] += pre[i - 1];
    if (pre[i] > 0) {
      dp[i] = 0;
    } else {
      if (real_mx == 0) {
        dp[i] = add(sum[i - 1], 1);
      } else {
        dp[i] = add(sum[i - 1], -sum[real_mx - 1]);
      }
    }
    sum[i] = add(dp[i], +sum[i - 1]);
    real_mx = max(real_mx, mx[i]);
  }
  if (real_mx != 0) {
    return add(sum[n], -sum[real_mx - 1]);
  } else {
    return add(sum[n], +1);
  }
}
int main() {
  scanf("%d %d %d", &n, &k, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d %d %d", &l[i], &r[i], &v[i]);
  }
  long long res = 1;
  for (int i = 0; i < k; i++) {
    long long tmp = work(i);
    res = res * tmp % mod;
  }
  printf("%lld\n", res);
}
