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;
pair<int, int> a[500004];
int n, m, k, i, mod, t, nr, la, y[500005], pw[500004], l[500004], r[500004],
    x[500004], f[500004], sf[500004], dp[500004][2], s[500004][2];
long long sol;
bool ok;
int main() {
  scanf("%d %d %d", &n, &k, &m);
  for (i = 1; i <= m; i++) scanf("%d %d %d", &l[i], &r[i], &x[i]);
  pw[0] = 1;
  mod = 998244353;
  for (i = 1; i <= n; i++) pw[i] = (pw[i - 1] * 2) % mod;
  sol = 1LL;
  for (t = 0; t < k; t++) {
    for (i = 1; i <= n; i++) f[i] = 0;
    nr = 0;
    for (i = 1; i <= m; i++) {
      y[i] = x[i] & (1 << t);
      if (y[i] == 0) {
        a[++nr].first = l[i];
        a[nr].second = r[i];
      } else {
        f[l[i]]++;
        f[r[i] + 1]--;
      }
    }
    for (i = 1; i <= n; i++) f[i] += f[i - 1];
    sort(a + 1, a + nr + 1);
    if (nr > 0) {
      la = 1;
      for (i = 2; i <= nr; i++) {
        if (a[i].first == a[la].first) continue;
        while ((la >= 1) && (a[la].second >= a[i].second)) la--;
        a[++la] = a[i];
      }
      nr = la;
    }
    sf[0] = 0;
    for (i = 1; i <= n; i++) {
      sf[i] = sf[i - 1];
      if (f[i] == 0) sf[i]++;
    }
    ok = true;
    for (i = 1; i <= nr; i++) {
      if (sf[a[i].first - 1] == sf[a[i].second]) {
        ok = false;
        break;
      }
    }
    if (ok == false) {
      printf("%d\n", 0);
      return 0;
    }
    for (i = 0; i <= n; i++) {
      dp[i][0] = 0;
      dp[i][1] = 0;
      s[i][0] = 0;
      s[i][1] = 0;
    }
    if (nr == 0) {
      sol = (1LL) * sol * pw[sf[n]] % mod;
      continue;
    }
    sol = (1LL) * sol * pw[sf[a[1].first - 1]] % mod *
          pw[sf[n] - sf[a[nr].second]] % mod;
    la = 1;
    if (a[1].first == a[1].second) {
      dp[a[1].first][0] = 1;
      dp[a[1].first][1] = 0;
      s[a[1].first][0] = 1;
      s[a[1].first][1] = 0;
      la++;
    } else {
      if (f[a[1].first] == 0) {
        dp[a[1].first][0] = 1;
        s[a[1].first][0] = 1;
      } else {
        dp[a[1].first][0] = 0;
        s[a[1].first][0] = 0;
      }
      dp[a[1].first][1] = 1;
      s[a[1].first][1] = 1;
    }
    for (i = (a[1].first + 1); i <= a[nr].second; i++) {
      if (f[i] == 0)
        dp[i][0] = (dp[i - 1][0] + dp[i - 1][1]) % mod;
      else
        dp[i][0] = 0;
      if (a[la].second == i) {
        dp[i][1] = s[i - 1][0] - s[a[la].first - 1][0];
        if (dp[i][1] < 0) dp[i][1] += mod;
        la++;
      } else {
        dp[i][1] = (dp[i - 1][0] + dp[i - 1][1]) % mod;
      }
      s[i][0] = s[i - 1][0] + dp[i][0];
      s[i][0] %= mod;
      s[i][1] = s[i - 1][1] + dp[i][1];
      s[i][1] %= mod;
    }
    sol =
        (1LL) * sol * ((0LL) + dp[a[nr].second][0] + dp[a[nr].second][1]) % mod;
  }
  printf("%I64d\n", sol);
  return 0;
}
