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)]
f = [i for i in range(N)]
def find(x):
    if x != f[x]:
        f[x] = find(f[x])
    return f[x]
def union(x, y):
    fx, fy = find(x), find(y)
    if fx == fy:    return False
    f[fx] = fy
    return True
   
E1 = []
for a, b, c in E:
    if c == 0:
        union(a, b)
    else:
        E1.append((a, b))

par = [find(x) for x in range(N)]
for a, b in E1:
    if par[a] == par[b]:
        print('No')
        exit()
C = len(set(par))

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")