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:
#!/usr/bin/env python3
n, m, q = [int(item) for item in input().split()]

class UnionFind:
    def __init__(self, n):
        self.par = [i for i in range(n)]
        self.size = [1] * n
        self.rank = [0] * n

    def find(self, x):
        if self.par[x] == x:
            return x
        else:
            self.par[x] = self.find(self.par[x])
            return self.par[x]

    def same_check(self, x, y):
        return self.find(x) == self.find(y)

    def get_size(self, x):
        return self.size[self.find(x)]

    def union(self, x, y):
        x = self.find(x)
        y = self.find(y)

        if self.rank[x] < self.rank[y]:
            self.par[x] = y
        else:
            self.par[y] = x
            if self.rank[x] == self.rank[y]:
                self.rank[x] += 1
    
Forest = UnionFind(n)
circles = []
group_num = n
edge_num = 0
seen = set()
for i in range(q):
    a, b, c = [int(item) for item in input().split()]
    if a > b:
        a, b = b, a
    if (a, b, c) in seen:
        continue
    if c == 0:
        # loop with simple path
        if not Forest.same_check(a, b):
            Forest.union(a, b)
            group_num -= 1
            edge_num += 1
    else:
        circles.append((a, b, c))
    seen.add((a, b, c))

for a, b, c in circles:
    # multipath in tree
    if Forest.same_check(a, b):
        print("No")
        exit()
    if group_num == 2:
        print("No")
        exit()

if len(circles) > 0:
    edge_min = edge_num + group_num
    edge_max = edge_num + group_num * (group_num - 1) // 2
else:
    edge_min = edge_num + group_num - 1
    edge_max = edge_num + group_num * (group_num - 1) // 2

if edge_min <= m <= edge_max:
    print("Yes")
else:
    print("No")