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 long long mod = 998244353;
int n, k, m;
struct st {
  int l, r, x;
};
st a[500005];
int b[500005];
int c[500005];
long long dp[500005][2];
long long s[500005];
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);
  }
  long long ans = 1;
  for (int j = 0; j < k; j++) {
    memset(b, 0, sizeof(b));
    memset(c, -1, sizeof(c));
    for (int i = 1; i <= m; i++) {
      if (a[i].x & (1 << j)) {
        b[a[i].l]++;
        b[a[i].r + 1]--;
      } else {
        c[a[i].r] = max(c[a[i].r], a[i].l);
      }
    }
    int x = -1;
    for (int i = 1; i <= n; i++) {
      if (c[i] != -1 && c[i] <= x) c[i] = -1;
      x = max(x, c[i]);
    }
    for (int i = 1; i <= n; i++) {
      b[i] = b[i - 1] + b[i];
    }
    memset(dp, 0, sizeof(dp));
    memset(s, 0, sizeof(s));
    dp[0][1] = 1;
    for (int i = 1; i <= n; i++) {
      if (c[i] != -1) {
        dp[i][0] = (dp[i - 1][0] + dp[i - 1][1]) % mod;
        dp[i][1] = (s[i - 1] - s[c[i] - 1] + mod) % mod;
      } else {
        dp[i][0] = (dp[i - 1][0] + dp[i - 1][1]) % mod;
        dp[i][1] = (dp[i - 1][0] + dp[i - 1][1]) % mod;
      }
      if (b[i] > 0) dp[i][0] = 0;
      s[i] = (s[i - 1] + dp[i][0]) % mod;
    }
    ans = ans * (dp[n][0] + dp[n][1]) % mod;
  }
  cout << ans << endl;
  return 0;
}
