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 N = 5e5 + 5;
const long long mod = 998244353;
long long add(long long a, long long b) {
  return (a + b >= mod ? a + b - mod : a + b);
}
long long sub(long long a, long long b) {
  return (a - b < 0 ? a - b + mod : a - b);
}
long long mul(long long a, long long b) { return (1LL * a * b) % mod; }
long long n, k, m, dp[N], diff[N], sum[N];
int check[N], lef[N];
pair<pair<int, int>, int> a[N];
bool cmp(pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
  if (a.first.first != b.first.first) {
    return a.first.first < b.first.first;
  } else {
    return a.first.second > b.first.second;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k >> m;
  for (int i = 1; i <= m; i++) {
    cin >> a[i].first.first >> a[i].first.second >> a[i].second;
  }
  sort(a + 1, a + 1 + m, cmp);
  long long ans = 1;
  for (int i = 0; i < k; i++) {
    for (int j = 1; j <= n; j++) {
      diff[j] = lef[j] = 0;
    }
    int mn = 1e9;
    for (int j = m; j >= 1; j--) {
      check[j] = 0;
      if ((a[j].second >> i) & 1) {
        check[j] = 1;
        continue;
      }
      if (mn <= a[j].first.second) {
        check[j] = 1;
      }
      mn = min(mn, a[j].first.second);
    }
    for (int j = 1; j <= m; j++) {
      if ((a[j].second >> i) & 1) {
        diff[a[j].first.first]++;
        diff[a[j].first.second + 1]--;
      } else {
        if (check[j] == 1) continue;
        lef[a[j].first.second] = a[j].first.first;
      }
    }
    dp[0] = sum[0] = 1;
    int preL = 0;
    for (int j = 1; j <= n + 1; j++) {
      diff[j] = diff[j - 1] + diff[j];
      if (j == n + 1) diff[j] = 0;
      if (diff[j] > 0) {
        dp[j] = 0;
      } else {
        if (preL == 0) {
          dp[j] = sum[j - 1];
        } else {
          dp[j] = sub(sum[j - 1], sum[preL - 1]);
        }
      }
      sum[j] = add(sum[j - 1], dp[j]);
      if (lef[j] > 0) preL = lef[j];
    }
    ans = mul(ans, dp[n + 1]);
  }
  cout << ans;
}
