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 = 500005;
const int MOD = 998244353;
int add(int x, int y) {
  x += y;
  return x >= MOD ? x - MOD : x;
}
int mul(int x, int y) { return 1ll * x * y % MOD; }
int sub(int x, int y) {
  x -= y;
  return x < 0 ? x + MOD : x;
}
int n, k, m, dp[N];
int L[N], R[N], X[N], a[N], pre[N];
int v[N];
int ans = 1;
void solve() {
  for (int i = 1; i <= n; i++) v[i] = 0, pre[i] = 0, dp[i] = 0;
  for (int i = 1; i <= m; i++) {
    if (a[i] == 1)
      pre[L[i]]++, pre[R[i] + 1]--;
    else
      v[R[i]] = max(v[R[i]], L[i]);
  }
  for (int i = 1; i <= n; i++) pre[i] = add(pre[i], pre[i - 1]);
  dp[0] = 1;
  int sum = 1, now = 0;
  for (int i = 1; i <= n; i++) {
    if (!pre[i]) dp[i] = sum, sum = add(sum, dp[i]);
    if (v[i] > 0)
      while (now < v[i]) sum = sub(sum, dp[now]), dp[now++] = 0;
  }
  int S = 0;
  for (int i = 0; i <= n; i++) S = add(S, dp[i]);
  ans = mul(ans, S);
}
int main() {
  scanf("%d%d%d", &n, &k, &m);
  for (int i = 1; i <= m; i++) scanf("%d%d%d", &L[i], &R[i], &X[i]);
  for (int bit = 0; bit < k; bit++) {
    for (int i = 1; i <= m; i++) a[i] = (X[i] >> bit & 1);
    solve();
  }
  printf("%d\n", ans);
}
