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>
#pragma GCC optimize(2)
using namespace std;
const int mod = 998244353;
struct node {
  int l;
  int r;
  friend bool operator<(const node &a, const node &b) {
    if (a.r == b.r) return a.l < b.l;
    return a.r < b.r;
  }
};
vector<node> g[40];
vector<node> v[40];
int n, k, m;
int pos, maxx;
long long dp[500005], sum[5000005], ret;
long long ans = 1;
bool cmp(const node &a, const node &b) {
  if (a.l == b.l) return a.r < b.r;
  return a.l < b.l;
}
void divide(int l, int r, int val) {
  int _ind = 1;
  while (val) {
    if (val & 1) {
      g[_ind].push_back((node){l, r});
    } else {
      v[_ind].push_back((node){l, r});
    }
    val >>= 1;
    _ind++;
  }
  for (int i = _ind; i <= k; i++) {
    v[i].push_back((node){l, r});
  }
}
void init(int _ind) {
  for (int i = 0; i <= n + 1; i++) {
    dp[i] = -1;
    sum[i] = 0;
  }
  dp[0] = sum[0] = 1;
  maxx = 0;
  pos = 0;
  for (int i = 0, sz = g[_ind].size(); i < sz; i++) {
    if (pos < g[_ind][i].l) pos = g[_ind][i].l;
    while (pos <= g[_ind][i].r) {
      dp[pos] = 0;
      pos++;
    }
  }
  pos = 0;
}
long long solve(int _ind) {
  init(_ind);
  dp[0] = 1;
  sum[0] = 1;
  int sz = v[_ind].size();
  long long ret;
  for (int i = 1; i <= n + 1; i++) {
    if (dp[i] == -1) {
      dp[i] = 0;
      while (pos < sz && v[_ind][pos].r < i) {
        maxx = max(v[_ind][pos].l, maxx);
        pos++;
      }
      if (maxx == 0)
        dp[i] = sum[i - 1];
      else
        dp[i] = (sum[i - 1] - sum[maxx - 1] + mod) % mod;
      ret = dp[i];
    }
    sum[i] = (sum[i - 1] + dp[i]) % mod;
  }
  return ret;
}
signed main() {
  scanf("%d %d %d", &n, &k, &m);
  for (int i = 1; i <= m; i++) {
    int l, r, x;
    scanf("%d %d %d", &l, &r, &x);
    divide(l, r, x);
  }
  for (int i = 1; i <= k; i++) {
    sort(v[i].begin(), v[i].end());
    sort(g[i].begin(), g[i].end(), cmp);
  }
  for (int i = 1; i <= k; i++) ans = ans * solve(i) % mod;
  cout << ans;
  return 0;
}
