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 long long mod = 998244353;
int n, m, k;
int p[40];
struct query {
  int l, r, x;
};
struct query q[500005];
int max0[500005], max1[500005];
long long dp[500005];
long long s[500005];
void solve() {
  cin >> n >> k >> m;
  p[0] = 1;
  for (int i = 1; i < 40; i++) p[i] = (p[i - 1] << 1);
  n++;
  for (int(i) = (1); (i) <= (m); ++(i)) cin >> q[i].l >> q[i].r >> q[i].x;
  m++;
  q[m].l = n;
  q[m].r = n;
  q[m].x = 0;
  long long ans = 1;
  for (int(j) = (0); (j) <= (k - 1); ++(j)) {
    for (int(i) = (0); (i) <= (n); ++(i)) {
      max0[i] = max1[i] = -1;
    }
    for (int(i) = (1); (i) <= (m); ++(i)) {
      if (q[i].x & p[j]) {
        if (max1[q[i].l] < q[i].r) max1[q[i].l] = q[i].r;
      } else {
        if (max0[q[i].r] < q[i].l) max0[q[i].r] = q[i].l;
      }
    }
    for (int i = 0; i <= n; i++) s[i] = 0;
    int L0 = 0;
    int R1 = -1;
    dp[0] = 1;
    s[0] = 1;
    for (int i = 1; i <= n; i++) {
      R1 = max(R1, max1[i]);
      if (i <= R1) {
        dp[i] = 0;
      } else {
        dp[i] = (L0 == 0 ? s[i - 1] : s[i - 1] - s[L0 - 1]);
        dp[i] %= mod;
      }
      s[i] = (s[i - 1] + dp[i]) % mod;
      L0 = max(L0, max0[i]);
    }
    ans = (ans * dp[n]) % mod;
  }
  if (ans < 0) ans += mod;
  cout << (ans) << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T = 1;
  for (int t = 1; t <= T; t++) solve();
}
