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 mod = 998244353;
const int maxn = 5e5;
int father[maxn + 5], mx[maxn + 5];
bool bo[maxn + 5];
int l[maxn + 5], r[maxn + 5], x[maxn + 5];
int n, m, k;
int f[maxn + 5];
int dp[maxn + 5], sum[maxn + 5];
int _find(int x) {
  if (x == father[x]) return x;
  father[x] = _find(father[x]);
  return father[x];
}
void _merge(int u, int v) {
  u = _find(u);
  v = _find(v);
  bo[v] = 1;
  if (u == v) return;
  father[u] = v;
  mx[v] = max(mx[v], mx[u]);
}
void cover(int L, int R) {
  int cur = L;
  while (cur <= R) {
    int nxt = mx[_find(cur)] + 1;
    _merge(cur, min(nxt, R));
    cur = nxt;
  }
}
int calc(int bt) {
  for (int i = 1; i <= n + 1; i++) {
    father[i] = i;
    mx[i] = i;
    bo[i] = 0;
    f[i] = 0;
  }
  for (int i = 1; i <= m; i++) {
    if (x[i] & (1 << bt))
      cover(l[i], r[i]);
    else {
      f[r[i] + 1] = max(f[r[i] + 1], l[i]);
    }
  }
  for (int i = 1; i <= n + 1; i++) bo[i] = bo[_find(i)];
  for (int i = 1; i <= n + 1; i++) f[i] = max(f[i], f[i - 1]);
  dp[0] = 1;
  sum[0] = 1;
  for (int i = 1; i <= n + 1; i++) {
    if (bo[i])
      dp[i] = 0;
    else if (f[i] >= 1)
      dp[i] = (sum[i - 1] - sum[f[i] - 1] + mod) % mod;
    else
      dp[i] = sum[i - 1];
    sum[i] = (sum[i - 1] + dp[i]) % mod;
  }
  return dp[n + 1];
}
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]);
  int ans = 1;
  for (int bt = 0; bt < k; bt++) ans = 1LL * ans * calc(bt) % mod;
  printf("%d\n", ans);
  return 0;
}
