1327_F. AND Segments
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].
```python
def main():
    import sys
    input = sys.stdin.buffer.readline

    mod = 998244353
    N, K, M = map(int, input().split())
    cond = []
    for _ in range(M):
       cond.append(tuple(map(int, input().split())))

    ans = 1
    for k in range(K):
        one = [0] * (N+1)
        zero_cond = [0] * (N+1)
        for l, r, x in cond:
            if x >> k & 1:
                one[l-1] += 1
                one[r] -= 1
            else:
                zero_cond[r] = max(zero_cond[r], l)
        for i in range(N):
            one[i+1] += one[i]
        dp = [0] * (N+1)
        dp[0] = 1
        cs = [0] * (N+2)
        cs[1] = 1
        R = 0
        for i in range(N):
            if not one[i]:
                dp[i+1] = (cs[i+1] - cs[R])%mod
            cs[i+2] = (cs[i+1] + dp[i+1])%mod
            R = max(R, zero_cond[i+1])
        ans = (ans * (cs[-1] - cs[R])%mod)%mod
    print(ans)


if __name__ == '__main__':
    main()

```