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 <cstdio>
#include <vector>
#include <iostream>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int MAXN = 100000;
vector<pii>vec;
int fa[MAXN + 5];
int find(int x) {
	return fa[x] = (fa[x] == x ? x : find(fa[x]));
}
void unite(int x, int y) {
	int fx = find(x), fy = find(y);
	if( fx != fy )
		fa[fx] = fy;
}
int N, Q; ll M;
int main() {
	scanf("%d%lld%d", &N, &M, &Q);
	for(int i=1;i<=N;i++) fa[i] = i;
	for(int i=1;i<=Q;i++) {
		int A, B, C; scanf("%d%d%d", &A, &B, &C), A++, B++;
		if( C == 0 ) unite(A, B);
		else vec.push_back(make_pair(A, B));
	}
	int cnt = 0;
	for(int i=1;i<=N;i++)
		if( find(i) == i ) cnt++;
	ll P = N - cnt, mx = 1LL*(cnt - 1)*cnt/2 + P;
	if( mx < M ) {
		puts("No");
		return 0;
	}
	ll mn = (vec.empty() ? N - 1 : N);
	if( mn > M ) {
		puts("No");
		return 0;
	}
	for(int i=0;i<vec.size();i++)
		if( find(vec[i].first) == find(vec[i].second) ) {
			puts("No");
			return 0;
		}
	puts("Yes");
}