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 sys
from collections import deque

input=sys.stdin.buffer.readline

n,m,k=map(int,input().split())
OUT=[[] for i in range(n)]
for i in range(m):
    u,v,w=map(int,input().split())
    OUT[u-1].append((w,v-1))

IN_ban=[0 for i in range(n)]
nth=[[] for i in range(10)]
for u in range(n):
    OUT[u].sort()
    nth[len(OUT[u])].append(u)
    for i in range(len(OUT[u])):
        w,v=OUT[u][i]
        id1,id2=len(OUT[u]),i
        IN_ban[v]|=1<<(id1*(id1-1)//2+id2)

use_cond=[[0 for i in range(10)] for j in range(10)]

for i in range(1,k+1):
    for j in range(i):
        check=[0]*n
        for u in nth[i]:
            w,v=OUT[u][j]
            if check[v]:
                break
            else:
                check[v]=1
                id1,id2=i,j
                use_cond[i][j]|=IN_ban[v]-(1<<(i*(i-1)//2+j))
        else:
            continue
        use_cond[i][j]=-1

deq=deque([(1,0,0)])
while deq and deq[0][0]<k+1:
    p,cond,used=deq.popleft()
    for i in range(p):
        if cond>>(p*(p-1)//2+i) &1==0 and use_cond[p][i]!=-1:
            tmp=used+(1<<(p*(p-1)//2+i))
            tmpcond=cond
            if tmp&use_cond[p][i]:
                continue
            deq.append((p+1,tmpcond|use_cond[p][i],tmp))

print(len(deq))
