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;
vector<pair<long long, long long> > op[500004];
long long prefx[500004];
long long dp[500004][2];
long long ison[500004];
long long mx[500004];
long long mod = 998244353;
long long take[500004];
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n, m, k;
  scanf("%d", &n);
  scanf("%d", &k);
  scanf("%d", &m);
  int i, j;
  for (i = 0; i < m; i++) {
    int x, y, z;
    scanf("%d", &x);
    scanf("%d", &y);
    scanf("%d", &z);
    op[x].push_back(make_pair(y, z));
  }
  for (i = 1; i <= n; i++) sort(op[i].begin(), op[i].end());
  long long ans = 1;
  for (i = 0; i < k; i++) {
    ison[n] = 0;
    long long now = (1 << i);
    for (j = 1; j <= n; j++) {
      prefx[j] = 0;
      dp[j][0] = 0;
      dp[j][1] = 0;
      mx[j] = 0;
      take[j] = 0;
      int c;
      bool ok = 0;
      for (c = 0; c < op[j].size(); c++) {
        if (op[j][c].second & now)
          ison[j]++, ison[op[j][c].first + 1]--;
        else if (!ok)
          mx[j] = op[j][c].first, ok = 1;
      }
    }
    long long mn = 1e18;
    for (j = n; j >= 1; j--) {
      if (mx[j] != 0 && mx[j] <= mn)
        mn = mx[j];
      else
        mx[j] = 0;
    }
    dp[0][1] = 1;
    dp[0][0] = 0;
    for (j = 1; j <= n; j++) {
      ison[j] += ison[j - 1];
      ison[j - 1] = 0;
      prefx[j] += prefx[j - 1];
      prefx[j] %= mod;
      prefx[j] += mod;
      prefx[j] %= mod;
      if (ison[j]) {
        dp[j][1] += (dp[j - 1][1] + dp[j - 1][0]);
        dp[j][1] %= mod;
        dp[j][1] += mod;
        dp[j][1] %= mod;
        dp[j][0] = 0;
      } else {
        dp[j][1] += (dp[j - 1][1] + dp[j - 1][0]);
        dp[j][1] %= mod;
        dp[j][1] += mod;
        dp[j][1] %= mod;
        dp[j][0] += (dp[j - 1][1] + dp[j - 1][0]);
        dp[j][0] %= mod;
        dp[j][0] += mod;
        dp[j][0] %= mod;
      }
      if (mx[j] != 0) {
        long long all = dp[j - 1][0] + dp[j - 1][1];
        all %= mod;
        all += mod;
        all %= mod;
        all -= prefx[j];
        all %= mod;
        all += mod;
        all %= mod;
        dp[mx[j]][1] -= all;
        dp[mx[j]][1] %= mod;
        dp[mx[j]][1] += mod;
        dp[mx[j]][1] %= mod;
        take[mx[j]] += all;
        take[mx[j]] %= mod;
        take[mx[j]] += mod;
        take[mx[j]] %= mod;
        prefx[j] += all;
        prefx[j] %= mod;
        prefx[j] += mod;
        prefx[j] %= mod;
      }
      prefx[j] -= take[j];
      prefx[j] %= mod;
      prefx[j] += mod;
      prefx[j] %= mod;
    }
    long long res = dp[n][1] + dp[n][0];
    res %= mod;
    res += mod;
    res %= mod;
    ans *= res;
    ans %= mod;
    ans += mod;
    ans %= mod;
  }
  printf("%I64d", ans);
  return 0;
}
