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<cmath>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<vector>
#define int long long
#define N 200010
using namespace std;
inline int read()
{
	int neg=1,num=0;char c=getchar();
	for(;!isdigit(c);c=getchar())if(c=='-')neg=-1;
	for(;isdigit(c);c=getchar())num=(num<<1)+(num<<3)+c-'0';
	return neg*num;
}
int n,m,Q,tot,fa[N],u[N],v[N],c[N]; 
inline int Find(int x){return fa[x]==x?x:fa[x]=Find(fa[x]);}
inline int Union(int x,int y)
{
	x=Find(x);y=Find(y);
	if(x==y)return 0;
	fa[y]=x;return 1;
}
signed main()
{
	n=read();m=read();Q=read();tot=n;
	for(int i=1;i<=n;i++)fa[i]=i;
	for(int i=1;i<=Q;i++)
	{
		u[i]=read()+1;v[i]=read()+1;c[i]=read();
		if(c[i])continue; 
		tot-=Union(u[i],v[i]);
	}
	if(1ll*tot*(tot-1)/2+n-tot<m)printf("No\n"),exit(0);//ÄÜÁ¬µÄ±ßÌ«ÉÙ 
	for(int i=1;i<=Q;i++)if(c[i])if(Find(u[i])==Find(v[i]))printf("No\n"),exit(0);//Á½¸ö¶àÌõÂ·¾¶ÔÚÍ¬Ò»¸öÁªÍ¨¿é 
	int sum=0;
	for(int i=1;i<=Q;i++)if(c[i])sum++;
	if(sum && m<n)printf("No\n"),exit(0);//ÒªÓÐ»· 
	printf("Yes\n");
	return 0;
}
/*
4 4 3
0 1 0
1 2 1
2 3 0

1 2 0
2 3 1
3 4 0

1 2
3 4
*/