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:
// FUCKED UP FUCKED UP FUCKED UP FUCKED UP FUCKED UP
#include <bits/stdc++.h>
using namespace std;

#pragma GCC optimize("O2,no-stack-protector,unroll-loops,fast-math")

#define F first
#define S second
#define pb push_back
#define SZ(x) (ll)(x.size())
#define all(x) x.begin(),x.end()
#define MP make_pair

typedef long long ll;
typedef long double ld;
typedef pair<ll,ll> pll;

//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const ll maxn=1e5+10, maxm=5e4+10, lg=21, mod=998244353, inf=1e18;

ll n,m,q,par[maxn],A[maxn],B[maxn],C[maxn],comp;
ll gpar(ll v){return par[v]==v ? v:par[v]=gpar(par[v]);}
void mrg(ll v,ll u){
	v=gpar(v), u=gpar(u);
	if(v==u) return;
	comp--;
	par[v]=u;
}

int main(){
	ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0);
	
	iota(par,par+maxn,0);
	cin>>n>>m>>q; comp=n;
	for(int i=0;i<q;i++){
		cin>>A[i]>>B[i]>>C[i];
		if(C[i] && m==n-1) return cout<<"No",0;
		if(!C[i]) mrg(A[i],B[i]);
	}
	if(m==n-1) return cout<<"Yes",0;
	for(int i=0;i<q;i++)if(C[i] && gpar(A[i])==gpar(B[i])) return cout<<"No",0;
	cout<<(m<=n-comp+comp*(comp-1)/2 ? "Yes":"No");
	
	return 0;
}