Boboniu has a directed graph with n vertices and m edges.

The out-degree of each vertex is at most k.

Each edge has an integer weight between 1 and m. No two edges have equal weights.

Boboniu likes to walk on the graph with some specific rules, which is represented by a tuple (c_1,c_2,…,c_k). If he now stands on a vertex u with out-degree i, then he will go to the next vertex by the edge with the c_i-th (1≤ c_i≤ i) smallest weight among all edges outgoing from u.

Now Boboniu asks you to calculate the number of tuples (c_1,c_2,…,c_k) such that

  * 1≤ c_i≤ i for all i (1≤ i≤ k). 
  * Starting from any vertex u, it is possible to go back to u in finite time by walking on the graph under the described rules. 

Input

The first line contains three integers n, m and k (2≤ n≤ 2⋅ 10^5, 2≤ m≤ min(2⋅ 10^5,n(n-1) ), 1≤ k≤ 9).

Each of the next m lines contains three integers u, v and w (1≤ u,v≤ n,u≠ v,1≤ w≤ m), denoting an edge from u to v with weight w. It is guaranteed that there are no self-loops or multiple edges and each vertex has at least one edge starting from itself.

It is guaranteed that the out-degree of each vertex is at most k and no two edges have equal weight.

Output

Print one integer: the number of tuples.

Examples

Input


4 6 3
4 2 1
1 2 2
2 4 3
4 1 4
4 3 5
3 1 6


Output


2


Input


5 5 1
1 4 1
5 1 2
2 5 3
4 3 4
3 2 5


Output


1


Input


6 13 4
3 5 1
2 5 2
6 3 3
1 4 4
2 6 5
5 3 6
4 1 7
4 3 8
5 2 9
4 2 10
2 1 11
6 1 12
4 6 13


Output


1

Note

For the first example, there are two tuples: (1,1,3) and (1,2,3). The blue edges in the picture denote the c_i-th smallest edges for each vertex, which Boboniu chooses to go through.

<image>

For the third example, there's only one tuple: (1,2,2,2).

<image>

The out-degree of vertex u means the number of edges outgoing from u.
import copy
from itertools import permutations
from collections import defaultdict
import sys
import io
import os
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline

def main():
    n, m, k = map(int, input().split())

    out = [[] for _ in range(n)]

    for _ in range(m):
        u, v, w = map(int, input().split())
        u -= 1
        v -= 1
        
        out[u].append((w, v))
        
    inc = [set() for _ in range(n)]

    forbidden = set()
    for u in range(n):
        for i, (_, v) in enumerate(sorted(out[u]), start=1):
            if (len(out[u]), i) in inc[v]:
                forbidden.add((len(out[u]), i))
            else:
                inc[v].add((len(out[u]), i))
        
    forbidden_follow = defaultdict(set)
    for v in range(n):
        for (c1, v1), (c2, v2) in permutations(inc[v], 2):
            if c1 < c2:
                forbidden_follow[(c1, v1)].add((c2, v2))
                
    total = 0
    cur_forbidden = set()
    def solve(cur_k):
        nonlocal k, total, forbidden, cur_forbidden
        if cur_k == k+1:
            total += 1
            return
        for val in range(1, cur_k+1):
            if (cur_k, val) in forbidden or (cur_k, val) in cur_forbidden:
                continue
            old = copy.copy(cur_forbidden)
            cur_forbidden.update(forbidden_follow[(cur_k, val)])
            solve(cur_k+1)
            cur_forbidden = old
            
    solve(1)
    print(total)
    
main()
