Problem: Snuke's mother gave Snuke an undirected graph consisting of N vertices numbered 0 to N-1 and M edges. This graph was connected and contained no parallel edges or self-loops.

One day, Snuke broke this graph. Fortunately, he remembered Q clues about the graph. The i-th clue (0 \leq i \leq Q-1) is represented as integers A_i,B_i,C_i and means the following:

* If C_i=0: there was exactly one simple path (a path that never visits the same vertex twice) from Vertex A_i to B_i.
* If C_i=1: there were two or more simple paths from Vertex A_i to B_i.



Snuke is not sure if his memory is correct, and worried whether there is a graph that matches these Q clues. Determine if there exists a graph that matches Snuke's memory.

Constraints

* 2 \leq N \leq 10^5
* N-1 \leq M \leq N \times (N-1)/2
* 1 \leq Q \leq 10^5
* 0 \leq A_i,B_i \leq N-1
* A_i \neq B_i
* 0 \leq C_i \leq 1
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N M Q
A_0 B_0 C_0
A_1 B_1 C_1
\vdots
A_{Q-1} B_{Q-1} C_{Q-1}


Output

If there exists a graph that matches Snuke's memory, print `Yes`; otherwise, print `No`.

Examples

Input

5 5 3
0 1 0
1 2 1
2 3 0


Output

Yes


Input

4 4 3
0 1 0
1 2 1
2 3 0


Output

No


Input

10 9 9
7 6 0
4 5 1
9 7 0
2 9 0
2 3 0
4 1 0
8 0 0
9 1 0
3 0 0


Output

No

Solution (in PYTHON3):

Solution:
s = input().split(" ")
n = int(s[0])
m = int(s[1])
q = int(s[2])
path_list =[]
rank = [0 for i in range(n)]
multi = []

for i in range(q):
    sp = input().split(" ")
    path = [int(pstr) for pstr in sp]
    path_list.append(path)

par = [i for i in range(n)]



#union-find
#xの木の根を得る
def root_find(x):
    if par[x] != x:
        par[x] = root_find(par[x])
        return par[x]
    else:
        return x

#併合する
def unite(x,y):
    rx = root_find(x)
    ry = root_find(y)

    if rx != ry:
        if rank[x] > rank[y]:
            par[y] = rx
        elif rank[x] == rank[y]:
            par[y] = rx
            rank[x] += 1
        else:
            par[x] = ry


for i in range(q):
    a = path_list[i][0]
    b = path_list[i][1]
    c = path_list[i][2]
    
    if c == 0 :
        pa = root_find(a)
        pb = root_find(b)
        if pa != pb:
            unite(pa,pb)
    else:
        multi.append((a,b))
        

"""
だめなじょうけん
・a,bが木であるはずなのに閉路がある
・ありうる辺の上限よりもMが大きい、もしくは辺の下限N-1よりもMが小さい
"""

for a,b in multi:
    if root_find(a) == root_find(b):
        print("No")
        exit(0)
        
k = 0
for i in range(n):
    if root_find(i) == i:
        k += 1



if len(multi) > 0:
    min_edge = n

else:
    min_edge = n - 1

max_edge = n + k * (k - 3) // 2

if min_edge <= m <= max_edge:
    print("Yes")
else:
    print("No")