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 CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
using LL = long long;
constexpr int maxn = 120000;
void no(const char* s){
	cout << s;
	exit(0);
}
int p[maxn], A[maxn], B[maxn], C[maxn];
int fp(int u){
	return u == p[u] ? u : p[u] = fp(p[u]);
}
int main(){
	ios::sync_with_stdio(false);
	LL N, M, Q;
	cin >> N >> M >> Q;
	for(int i = 1; i <= Q; i += 1) cin >> A[i] >> B[i] >> C[i];
	if(M == N - 1){
		for(int i = 1; i <= Q; i += 1) if(C[i]) no("No");
		no("Yes");
	}
	for(int i = 1; i <= N; i += 1) p[i] = i;
	LL cc = N; 
	for(int i = 1; i <= Q; i += 1) if(not C[i]){
		int u = fp(A[i]), v = fp(B[i]);
		if(u != v) cc -= 1;
		p[u] = v;
	}
	if(cc == 2) no("No");
	for(int i = 1; i <= Q; i += 1) if(C[i])
		if(fp(A[i]) == fp(B[i])) no("No");
	if(M <= N - 1 + cc * (cc - 1) / 2 - cc + 1) no("Yes");
	no("No");
}