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;
template <class T>
inline void amin(T &x, const T &y) {
  if (y < x) x = y;
}
template <class T>
inline void amax(T &x, const T &y) {
  if (x < y) x = y;
}
using ll = signed long long;
using PII = pair<int, int>;
using VI = vector<int>;
using VVI = vector<VI>;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int MAXN = 700;
const int N = 500005;
int mod = 998244353;
int n, k, m;
int l[N], r[N], x[N];
int cnt[N], mx[N], dp[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  scanf("%d%d%d", &n, &k, &m);
  for (int i = (0); i < (m); i++) scanf("%d%d%d", l + i, r + i, x + i);
  ll ans = 1;
  for (int bit = (0); bit < (k); bit++) {
    for (int i = (0); i < (n + 1); i++) cnt[i] = 0, mx[i] = -1;
    for (int i = (0); i < (m); i++) {
      if ((x[i] >> bit) & 1) {
        cnt[l[i]]++;
        cnt[r[i] + 1]--;
      } else {
        mx[r[i]] = max(mx[r[i]], l[i]);
      }
    }
    for (int i = (1); i < (n + 1); i++) cnt[i] += cnt[i - 1];
    for (int i = (0); i < (n + 1); i++) dp[i] = 0;
    dp[0] = 1;
    ll sum = 1;
    int j = 0;
    for (int i = (1); i < (n + 1); i++) {
      if (cnt[i]) {
        dp[i] = 0;
      } else {
        dp[i] = sum;
        sum = sum * 2 % mod;
      }
      while (j < mx[i]) {
        sum -= dp[j++];
        sum %= mod;
        sum += mod;
        sum %= mod;
      }
    }
    ll res = 0;
    for (int i = (j); i < (n + 1); i++) res = (res + dp[i]) % mod;
    ans = ans * res % mod;
  }
  printf("%lld\n", ans);
  return 0;
}
