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;
#define int long long
#define rep(i,n) for(int i=0;i<n;++i)
typedef pair<int, int> P;
int par[100006], siz[100006];
void init(int n) {
	rep(i, n) {
		par[i] = i; siz[i] = 1;
	}
}
int find(int x) {
	if (par[x] == x)return x;
	return par[x] = find(par[x]);
}
void unite(int x, int y) {
	x = find(x), y = find(y);
	if (x == y)return;
	if (siz[x] < siz[y])swap(x, y);
	par[y] = x;
	siz[x] += siz[y];
}

vector<P>v1;
signed main() {
	int n, m, q; cin >> n >> m >> q;
	init(n);
	rep(i, q) {
		int a, b, c; cin >> a >> b >> c;
		if (c)v1.push_back(P(a, b));
		else unite(a, b);
	}

	if (v1.size()) {
		set<int>S;
		rep(i, n)S.insert(find(i));
		if (S.size() <= 2) {
			puts("No"); return 0;
		}
		rep(i, v1.size()) {
			if (find(v1[i].first) == find(v1[i].second)) {
				puts("No"); return 0;
			}
		}
		int M = 0;
		for (auto itr = S.begin(); itr != S.end(); itr++) {
			M += siz[find(*itr)] - 1;
		}
		int z = S.size();
		M += z*(z- 1) / 2;
		if(n-1<m&&m<=M)puts("Yes");
		else puts("No");
	}
	else {
		if(m==n-1)puts("Yes");
		else puts("No");
	}
}