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>
const int MAXN = 6e5 + 10;
const double eps = 1e-8;
const int inf = 998244353;
using namespace std;
struct edge {
  int t, v;
  edge *next;
} e[MAXN << 1], *h[MAXN], *o = e;
void add(int x, int y, int vul) {
  o->t = y;
  o->v = vul;
  o->next = h[x];
  h[x] = o++;
}
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return x * f;
}
int n, k, m;
typedef struct node {
  int l, r, x;
} node;
node d[MAXN];
int maxx[MAXN], a[MAXN], pos[MAXN];
long long sum[MAXN];
int main() {
  n = read();
  k = read();
  m = read();
  long long ans = 1;
  for (int i = 1; i <= m; i++)
    d[i].l = read(), d[i].r = read(), d[i].x = read();
  for (int i = 0; i <= k - 1; i++) {
    for (int j = 1; j <= n; j++) a[j] = maxx[j] = pos[j] = sum[j] = 0;
    for (int j = 1; j <= m; j++) {
      if ((d[j].x >> i) & 1)
        a[d[j].l]++, a[d[j].r + 1]--;
      else
        maxx[d[j].r] = max(maxx[d[j].r], d[j].l);
    }
    for (int j = 1; j <= n; j++) a[j] += a[j - 1];
    int id = 0;
    sum[0] = 1;
    bool flag = 0;
    for (int j = 1; j <= n; j++) {
      if (a[j] > 0)
        sum[j] = sum[j - 1];
      else
        sum[j] = (sum[j - 1] +
                  (sum[j - 1] - (id > 0 ? sum[id - 1] : 0) + inf) % inf) %
                 inf,
        flag = 1;
      id = max(id, maxx[j]);
    }
    if (id > 0) sum[n] = (sum[n] - sum[id - 1] + inf) % inf;
    ans = ans * sum[n] % inf;
  }
  printf("%lld\n", ans);
  return 0;
}
