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 = sys.stdin.buffer.readline
 
n,m,k = map(int,input().split())
edge = [[] for _ in range(n)]
 
dg = 10**6
MOD = 10**9+7
 
for _ in range(m):
    u,v,w = map(int,input().split())
    edge[u-1].append(w*dg + v-1)
 
num = [[] for i in range(k+1)]
 
one = set()
 
for i in range(n):
    num[len(edge[i])].append(i)
    edge[i].sort()
 
q = [[0]*(k+1) for i in range(k+1)]
q2 = [[0]*(k+1) for i in range(k+1)]

 
for i in range(1,k+1):
    for j in range(i):
        for e in num[i]:
            go = edge[e][j]%dg
            q[i][j] = (q[i][j] + ((go*go*go%MOD + go +122312)%MOD)*1213 + 12316)%MOD
            q2[i][j] = (q2[i][j] + (go*go+12231312)*1213 + 12316)%MOD
 
cor = 0
cor2 = 0
for i in range(n):
    cor = (cor + ((i*i*i%MOD + i +122312)%MOD)*1213 + 12316)%MOD 
    cor2 = (cor2 + (i*i+12231312)*1213 + 12316)%MOD

fac = 1
for i in range(2,k+1):
    fac *= i
 
 
 
res = 0
for w in range(1,fac+1):
 
    tmp = w
    nxt = 0
    nxt2 = 0
    tank = []
    for r in range(k,0,-1):
        tmp,c = divmod(tmp,r)
        tank.append(c)
        nxt = (nxt + q[r][c])%MOD
        nxt2 = (nxt2 + q2[r][c])%MOD
    if nxt == cor and nxt2 == cor2:

        res += 1
        
 
print(res)