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;
class trio {
 public:
  long long first, second, third;
};
int32_t main() {
  long long n, k, m;
  scanf("%lld", &n);
  scanf("%lld", &k);
  scanf("%lld", &m);
  trio arr[m];
  for (long long i = 0; i < m; ++i) {
    scanf("%lld", &arr[i].first);
    scanf("%lld", &arr[i].second);
    scanf("%lld", &arr[i].third);
  }
  long long ans = 1;
  for (long long i = 0; i < k; ++i) {
    long long brr[n + 2], crr[n + 2], dp[n + 2], dpsum[n + 2];
    memset(brr, 0, sizeof(brr));
    memset(crr, 0, sizeof(crr));
    memset(dp, 0, sizeof(dp));
    memset(dpsum, 0, sizeof(dpsum));
    for (long long j = 0; j < m; ++j) {
      if (arr[j].third & (1 << i)) {
        brr[arr[j].first]++;
        brr[arr[j].second + 1]--;
      } else {
        crr[arr[j].second] = max(arr[j].first, crr[arr[j].second]);
      }
    }
    for (long long i = 1; i < n + 2; ++i) {
      brr[i] += brr[i - 1];
    }
    long long maxi = 0;
    dp[0] = 1;
    dpsum[0] = 1;
    for (long long j = 1; j < n + 2; ++j) {
      if (brr[j]) {
        dp[j] = 0;
        dpsum[j] = dpsum[j - 1];
      } else {
        if (maxi == 0) {
          dp[j] = dpsum[j - 1];
        } else {
          dp[j] = dpsum[j - 1] - dpsum[maxi - 1] + 998244353;
        }
        dpsum[j] = dpsum[j - 1] + dp[j];
      }
      dp[j] %= 998244353;
      dpsum[j] %= 998244353;
      maxi = max(maxi, crr[j]);
    }
    ans *= (dp[n + 1]) % 998244353;
    ans %= 998244353;
  }
  printf("%lld ", ans);
  return 0;
}
