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:
class UnionFind:
    def __init__(self, n):
        self.par = [-1 for i in range(n+1)]
        self.rank = [0] * (n+1)
 
    # 検索
    def find(self, x):
        if self.par[x] < 0:
            return x
        else:
            self.par[x] = self.find(self.par[x])
            return self.par[x]
 
    # 併合
    def union(self, x, y):
        x = self.find(x)
        y = self.find(y)
        if not x==y: # 根が同じでない場合のみ併合する
            if self.rank[x] < self.rank[y]:
                self.par[y] += self.par[x] # 要素数を併合
                self.par[x] = y # 根を付け替えている
            else:
                self.par[x] += self.par[y]
                self.par[y] = x
                if self.rank[x] == self.rank[y]:
                    self.rank[x] += 1
        
    # 同じ集合に属するか判定
    def same_check(self, x, y):
        return self.find(x) == self.find(y) 
 
def main():
    import sys
    input = sys.stdin.readline
    
    N,M,Q = map(int,input().split())
    KI = UnionFind(N-1)
    CC = []
    for i in range(Q):
        A,B,C = map(int,input().split())
        if C == 0:
            KI.union(A,B)
        else:
            CC.append([A,B])
    
    flag = 0
    
    oya = {}
    for i in range(N):
        p = KI.find(i)
        oya[p]=1
    PPP = len(oya)
    
    for i in CC:
        A,B = i
        if KI.same_check(A,B):
            flag = 1
            break
        elif PPP<=2:
            flag = 1
            break
        elif M == N-1:
            flag = 1
            break
    
    if N-1+((PPP-2)*(PPP-1))//2 < M:
        flag = 1
            
    if flag == 0:
        print("Yes")
    else:
        print("No")
    
    
    
    
main()