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<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<vector>
using namespace std;
#define int long long
#define MAXN 200006
#define pb push_back
#define pii pair<int,int>
#define mp make_pair
int n , m , q;

int fa[MAXN];
int find( int x ) {
	return x == fa[x] ? x : fa[x] = find( fa[x] );
}
struct dl{
	int u , v;
} d[MAXN] ;
int cnt , c;

signed main( ) {
	cin >> n >> m >> q;
	for( int i = 1 ; i <= n ; ++ i ) fa[i] = i;
	for( int i = 1 , u , v , c ; i <= q ; ++ i ) {
		scanf("%lld%lld%lld",&u,&v,&c);
		++ u , ++ v;
		if( !c ) {
			fa[find( u )] = find( v );
		} else {
			d[++ cnt] = (dl) { u , v };
		}
	}
	for( int i = 1 , u , v ; i <= cnt ; ++ i ) {
		if( find( d[i].u ) == find( d[i].v ) )
			return puts("No") , 0;
	}
	for( int i = 1 ; i <= n ; ++ i ) if( fa[i] == i )
		++ c;
	int ed = m - n + c;
	if( !cnt ) return puts( ( ed >= c - 1 && ed <= c * ( c - 1 ) / 2 ) ? "Yes" : "No" ) , 0;
	else return puts( ( ed >= c && ed <= c * ( c - 1 ) / 2 ) ? "Yes" : "No" ) , 0;
}