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;
int n, m, k;
const long long mod = 998244353;
struct node {
  int l, r, x;
} infor[500010];
int one[500010], mx[500010];
long long sum[500010], ans = 1;
int main() {
  scanf("%d%d%d", &n, &k, &m);
  for (int i = 0; i < m; i++) {
    int l, r, x;
    scanf("%d%d%d", &infor[i].l, &infor[i].r, &infor[i].x);
  }
  for (int i = 0; i < k; i++) {
    memset(one, 0, sizeof(one));
    memset(mx, 0, sizeof(mx));
    memset(sum, 0, sizeof(sum));
    sum[0] = 1;
    for (int j = 0; j < m; j++) {
      if (infor[j].x & (1 << i)) {
        one[infor[j].l]++;
        one[infor[j].r + 1]--;
      } else {
        mx[infor[j].r] = max(mx[infor[j].r], infor[j].l);
      }
    }
    for (int i = 1; i <= n; i++) {
      one[i] += one[i - 1];
    }
    int p = 0;
    long long ret = 1;
    for (int j = 1; j <= n; j++) {
      if (!one[j]) {
        sum[j] = ret;
        ret *= 2;
        ret %= mod;
      }
      while (p < mx[j]) {
        ret = (ret - sum[p] + mod) % mod;
        p++;
      }
    }
    ans *= ret;
    ans %= mod;
  }
  cout << ans << endl;
  return 0;
}
