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<cstdio>
#include<algorithm>
using namespace std;
int n,Q;
long long m;
#define Maxn 100010
int head[Maxn],v[Maxn<<1],nxt[Maxn<<1],tot=0;
int bel[Maxn],col=0;
inline void add_edge(int s,int e){
	tot++;v[tot]=e;nxt[tot]=head[s];head[s]=tot;
	tot++;v[tot]=s;nxt[tot]=head[e];head[e]=tot;
}
void dfs(int u){
	bel[u]=col;
	for(int i=head[u];i;i=nxt[i])
	    if(!bel[v[i]])dfs(v[i]);
}

struct P{
	int a,b;
}seq[Maxn];
int len=0;

int main(){
	scanf("%d%lld%d",&n,&m,&Q);
	int a,b,c;
	for(register int i=1;i<=Q;++i){
		scanf("%d%d%d",&a,&b,&c);
		a++;b++;
		if(!c)add_edge(a,b);
		else seq[++len]=(P){a,b};
	}
	for(register int i=1;i<=n;++i)
	    if(!bel[i]){
			col++;
			dfs(i);
	    }
	for(register int i=1;i<=len;++i)
	    if(bel[seq[i].a]==bel[seq[i].b]){
			puts("No");
			return 0;
	    }
	m-=n-(col);
	if(m<0){
		puts("No");
		return 0;
	}
	if(len){
		if(m>=col&&m<=1ll*col*(col-1)/2)puts("Yes");
		else puts("No");
	}else{
		if(m<=1ll*col*(col-1)/2&&m>=col-1)puts("Yes");
		else puts("No");
	}
	return 0;
}