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:
import sys
input=sys.stdin.readline

def find_parent(x):
    y=parent[x]
    if y<0:
        return x
    parent[x]=find_parent(y)
    return parent[x]

def connect(a,b):
    c=find_parent(a)
    d=find_parent(b)
    if c==d:
        return
    if parent[c]<parent[d]:
        parent[c]+=parent[d]
        parent[d]=c
    else:
        parent[d]+=parent[c]
        parent[c]=d
    return

N,M,Q=map(int,input().split())
data=[[],[]]
for i in range(Q):
    a,b,c=map(int,input().split())
    data[c].append((a,b))
    
parent=[-1]*N
for a,b in data[0]:
    connect(a,b)

cnt=0
for u in parent:
    if u<0:
        cnt+=1

for a,b in data[1]:
    if find_parent(a)==find_parent(b):
        print("No")
        sys.exit()

if M==N-1:
    if not data[1]:
        print("Yes")
    else:
        print("No")
else:
    M-=N
    if M<=cnt*(cnt-1)//2-cnt:
        print("Yes")
    else:
        print("No")