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<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
using namespace std;
#define MAXN 100000
#define LL long long
LL read(){
    LL x=0,F=1;char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')F=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*F;
}
struct ck{
	int u,v;
}p[MAXN+5];
int n,k,cnt,tot;
int fa[MAXN+5],siz[MAXN+5];
LL m,sum;
vector<int> G[MAXN+5];
int xfind(int x){
	return (fa[x]==x)?x:fa[x]=xfind(fa[x]);
}
int main()
{
	n=read(),m=read(),k=read();
	for(int i=1;i<=n;i++)fa[i]=i,siz[i]=1;
	for(int i=1;i<=k;i++){
		int u=read()+1,v=read()+1,opt=read();
		if(opt==0){
			int a=xfind(u),b=xfind(v);
			if(a!=b){
				fa[b]=a;siz[a]+=siz[b];
			}
		}else p[++cnt]=(ck){u,v};
	}
	for(int i=1;i<=cnt;i++)
	if(xfind(p[i].u)==xfind(p[i].v))
	{printf("No\n");return 0;}
	for(int i=1;i<=n;i++)
	if(xfind(i)==i)tot++;
	if(tot<=2&&cnt){printf("No\n");return 0;}
	if(n-1==m&&cnt){printf("No\n");return 0;}
	if(1LL*tot*(tot-1)/2+n-tot>=m){printf("Yes\n");return 0;}
	else {printf("No\n");return 0;}
}