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;
typedef long long ll;
const int INF = (1<<30);
const ll INFLL = (1ll<<60);
const ll MOD = (ll)(1e9+7);

#define l_ength size

void mul_mod(ll& a, ll b){
	a *= b;
	a %= MOD;
}

void add_mod(ll& a, ll b){
	a = (a<MOD)?a:(a-MOD);
	b = (b<MOD)?b:(b-MOD);
	a += b;
	a = (a<MOD)?a:(a-MOD);
}

int a[100100],b[100100],c[100100],ufp[100100],ufr[100100];

void ufinit(int n){
	int i;
	for(i=0; i<n; ++i){
		ufp[i] = i;
		ufr[i] = 1;
	}
}

int ufind(int x){
	if(x==ufp[x]){
		return x;
	}
	return ufp[x] = ufind(ufp[x]);
}

void unionf(int x, int y){
	x = ufind(x);
	y = ufind(y);
	if(x==y){
		return;
	}
	if(ufr[x] < ufr[y]){
		swap(x,y);
	}
	ufr[x] += ufr[y];
	ufp[y] = x;
}

int main(void){
	int n,q,i;
	ll m,k=0ll;
	cin >> n >> m >> q;
	ufinit(n);
	for(i=0; i<q; ++i){
		cin >> a[i] >> b[i] >> c[i];
		if(!c[i]){
			unionf(a[i],b[i]);
		}
		if(n>m && c[i]){
			cout << "No" << endl;
			return 0;
		}
	}
	if(n>m){
		cout << "Yes" << endl;
		return 0;
	}
	for(i=0; i<q; ++i){
		if(c[i] && ufind(a[i])==ufind(b[i])){
			cout << "No" << endl;
			return 0;
		}
	}
	for(i=0; i<n; ++i){
		if(ufind(i)==i){
			++k;
		}
	}
	cout << ((n<=m && m<=(n+k*(k-3)/2))?"Yes":"No") << endl;
	return 0;
}