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 rep( i, s, t ) for( register int i = s; i <= t; ++ i )
#define re register
#define int long long
int gi() {
	char cc = getchar() ; int cn = 0, flus = 1 ;
	while( cc < '0' || cc > '9' ) {  if( cc == '-' ) flus = - flus ; cc = getchar() ; }
	while( cc >= '0' && cc <= '9' )  cn = cn * 10 + cc - '0', cc = getchar() ;
	return cn * flus ;
}
const int N = 1e5 + 5 ; 
int n, m, T, num, cnt, idnex, fa[N] ; 
struct Q {
	int a, b, c ; 
} q[N] ;
int fd( int x ) {
	return ( x == fa[x] ) ? x : fa[x] = fd(fa[x]) ;
}
void merge( int x, int y ) {
	int u = fd(x), v = fd(y) ; 
	if( u != v ) -- num, fa[u] = v, ++ cnt ; 
}
signed main()
{
	n = gi(), m = gi(), T = gi() ; 
	rep( i, 1, n ) fa[i] = i ; num = n ;
	rep( i, 1, T ) {
		q[i].a = gi(), q[i].b = gi(), q[i].c = gi() ;
		if( q[i].c == 0 ) ++ idnex, merge( q[i].a, q[i].b ) ; 
	}
	if( m <= n - 1 ) {
		if( m < n - 1 ) puts("No") ;
		else if( T == idnex ) puts("Yes") ;
		else puts("No") ;
		exit(0) ;
	}
	int l = n, r = cnt + ( num * ( num - 1 ) ) / 2 ;
	rep( i, 1, T ) {
		if( q[i].c == 1 )
		if( fd(q[i].a) == fd(q[i].b) ) {
			puts("No") ; exit(0) ;
		}
	}
	if( m >= l && m <= r ) puts("Yes") ;
	else puts("No") ;
	return 0 ;
}