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 inf = 1e9 + 7;
const int max_n = 5e5 + 1, max_m = 5e5 + 1, max_k = 30;
int n, m, k;
int after[max_k][max_n], num[max_k][max_n];
int mod = 998244353;
int le[max_k][max_n];
void solve1() {
  cin >> n >> k >> m;
  for (int i = 0; i < k; i++) {
    for (int j = 0; j < n; j++) {
      num[i][j] = -1;
      after[i][j] = j + 1;
      le[i][j] = -1;
    }
  }
  for (int q = 0; q < m; q++) {
    int l, r, temp1;
    cin >> l >> r >> temp1;
    l--;
    r--;
    for (int w = 0; w < k; w++) {
      if ((temp1 & (1 << w)) == 0) {
        le[w][r] = max(le[w][r], l);
        continue;
      }
      int v = l;
      while (v <= r) {
        num[w][v] = 1;
        int to = after[w][v];
        after[w][v] = after[w][r];
        v = to;
      }
    }
  }
}
long long dp[max_n];
long long get(int id) {
  if (id == -1) return 1;
  return dp[id];
}
long long solve2(int i) {
  int maxim = 0;
  long long sum = 0;
  if (num[i][0] == -1) sum = 1;
  for (int j = 0; j < n; j++) {
    if (num[i][j] == 1 && le[i][j] == j) {
      dp[j] = 0;
    } else if (le[i][j] == -1) {
      dp[j] = get(j - 1) % mod;
      if (num[i][j] == -1) dp[j] = 2 * dp[j] % mod;
    } else {
      while (maxim < le[i][j]) {
        if (num[i][maxim] == -1) sum -= get(maxim - 1);
        maxim++;
      }
      dp[j] = sum;
      dp[j] %= mod;
    }
    if (j + 1 == n || num[i][j + 1] == -1) sum += dp[j];
  }
  return dp[n - 1];
}
void solve() {
  solve1();
  int res = 1;
  for (int i = 0; i < k; i++) res = res * solve2(i) % mod;
  cout << res;
}
signed main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(0);
  solve();
}
