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 F first 
#define S second 
#define pii pair<int,int> 
#define mpr make_pair

typedef long long ll;

const int maxn = 1e5+10;
const int mod = 998244353;
const int inf = 1e9+10;

int n, k, m, q;
int a[maxn];
vector<pii> edge;
vector<int> g[maxn];
bool mark[maxn];
int comp[maxn];

void dfs(int v, int k)
{
	comp[v] = k;
	mark[v] = 1;
	for(auto u : g[v])
		if(!mark[u]) dfs(u,k);
}

signed main()
{
//	ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);	
	
	cin>> n >> m >> q;
	bool is_cyc = 0;
	for(int i = 1, a, b, c; i <= q; i++)
	{
		cin>> a >> b >> c;
		a++; b++;
		if(c == 0)
		{
			g[a].push_back(b);
			g[b].push_back(a);
		}
		else 
		{
			is_cyc = 1;
			edge.push_back({a,b});	
		}
	}	
	
	for(int i = 1; i <= n; i++) 
		if(!mark[i]) dfs(i,++k);
	
	for(auto e : edge)
	{
		int u = e.F, v = e.S;
		if(comp[u] == comp[v]) return cout<<"No", 0;
	}
	
	if(k <= 2 && is_cyc) return cout<<"No", 0;
	
	int L = n-1 + is_cyc, R = k * (k-1) / 2 + n-k;	
	if(m >= L && m <= R) cout<<"Yes";
	else cout<<"No";
}