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;
long long gcd(long long a, long long b) {
  for (; b; a %= b, swap(a, b))
    ;
  return a;
}
const long long MOD = 998244353;
int n, k, m;
struct query {
  int l, r, x;
} ql[500010], qr[500010];
bool c1(query q1, query q2) { return q1.l < q2.l; }
bool c2(query q1, query q2) { return q1.r < q2.r; }
long long dp[500010];
long long psum[500010];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  cin >> n >> k >> m;
  for (int i = 0; i < m; i++) {
    cin >> ql[i].l >> ql[i].r >> ql[i].x;
    qr[i] = ql[i];
  }
  sort(ql, ql + m, c1);
  sort(qr, qr + m, c2);
  long long ans = 1;
  for (int x = 0; x < k; x++) {
    memset(dp, 0, sizeof dp);
    memset(psum, 0, sizeof psum);
    int ql_ptr = 0;
    int qr_ptr = 0;
    int last0 = 0;
    int last1 = 0;
    for (int i = 1; i <= n + 1; i++) {
      while (ql_ptr < m && ql[ql_ptr].l == i) {
        if (ql[ql_ptr].x & (1 << x)) last1 = max(last1, ql[ql_ptr].r);
        ql_ptr++;
      }
      if (last1 == 0) {
        if (last0 == 0)
          dp[i] = (psum[i - 1] + 1) % MOD;
        else
          dp[i] = (psum[i - 1] - psum[last0 - 1] + MOD) % MOD;
      }
      psum[i] = (psum[i - 1] + dp[i]) % MOD;
      while (qr_ptr < m && qr[qr_ptr].r == i) {
        if (!(qr[qr_ptr].x & (1 << x))) last0 = max(last0, qr[qr_ptr].l);
        qr_ptr++;
      }
      while (last1 == i) last1 = 0;
      ;
    }
    ans *= dp[n + 1];
    ans %= MOD;
  }
  cout << ans;
}
