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].
import io, os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
from itertools import accumulate

n,k,m=map(int,input().split())
Q=[tuple(map(int,input().split())) for i in range(m)]

mod=998244353
S=1

for keta in range(k):
    ANS=[0]*(n+5)
    QK=[n+3]*(n+5)

    for l,r,x in Q:
        l-=1
        r-=1
        if x & (1<<keta)!=0:
            ANS[l]+=1
            ANS[r+1]-=1

        else:
            QK[l]=min(QK[l],r+1)

    ANS=tuple(accumulate(ANS))
    for i in range(n+3,-1,-1):
        QK[i]=min(QK[i],QK[i+1])

    PLUS=[0]*(n+5)

    PLUS[0]=1
    PLUS[1]=-1

    for i in range(n+1):
        PLUS[i]=(PLUS[i]+PLUS[i-1])%mod
            
        if ANS[i-1]==0:
            last=QK[i]

            PLUS[i+1]=(PLUS[i]+PLUS[i+1])%mod
            PLUS[last+1]=(PLUS[last+1]-PLUS[i])%mod

    S=S*(PLUS[n]+PLUS[n+1])%mod

print(S)
            
        
        
            
            
    
