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
input = lambda: sys.stdin.readline().rstrip()
N, M, K = map(int, input().split())
IN = [[] for _ in range(N)]
OUT = [[] for _ in range(N)]
t, m = 12345, (1 << 60) - 1
HASH = []
s = 987654321
H = 0
for _ in range(N):
    s = (s * 17 + t) & m
    HASH.append(s)
    H ^= s
for _ in range(M):
    u, v, w = map(int, input().split())
    u, v = u-1, v-1
    OUT[u].append((v, w))

X = [[0] * (i+2) for i in range(K)]
for i in range(N):
    c = len(OUT[i])
    for j, (v, w) in enumerate(sorted(OUT[i], key = lambda x: x[1])):
        X[c-1][j] ^= HASH[v]

s = 0
for i in range(K):
    s ^= X[i][0]
X[0][0] = 0

ans = 0
A = [0] * K
f = 1
while f:
    if s == H:
        ans += 1
    
    A[0] += 1
    a = 0
    while A[a] == a + 1:
        s ^= X[a][0]
        A[a] = 0
        a += 1
        if a >= K:
            f = 0
            break
        s ^= X[a][A[a]] ^ X[a][A[a]+1]
        A[a] += 1

print(ans)