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

N,M,Q=map(int,input().split())
E=[list(map(int,input().split())) for i in range(Q)]

# UnionFind

Group=[i for i in range(N)]# グループ分け.Group[i]=jのときiとjは同じグループ

def find(x):# find(a)=find(b)のとき同じグループ
    while Group[x] != x:
        x=Group[x]
    return x

def Union(x,y):  # xとyが同じグループになるよう更新
    if find(x) != find(y):
        Group[find(y)]=Group[find(x)]=min(find(y),find(x))

E1=[]
for a,b,x in E:
    if x==0:
        Union(a,b)
    else:
        E1.append((a,b))

Group=[find(i) for i in range(N)]

for a,b in E1:
    if Group[a]==Group[b]:
        print("No")
        sys.exit()

C=len(set(Group))

if len(E1)==0:
    if N-1<=M<=(N-C)+C*(C-1)//2:
        print("Yes")
    else:
        print("No")

else:
    if N<=M<=(N-C)+C*(C-1)//2:
        print("Yes")
    else:
        print("No")