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<set>
#include<map>
#include<stack>
#include<ctime>
#include<cmath>
#include<queue>
#include<cstdio>
#include<vector>
#include<climits>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define LL long long
using namespace std;
LL read(){
	bool f=0;LL x=0;char c=getchar();
	while(c<'0'||'9'<c){if(c=='-')f=1;c=getchar();}
	while('0'<=c&&c<='9') x=(x<<3)+(x<<1)+(c^48),c=getchar();
	return !f?x:-x;
}
#define MAXN 100000
#define INF 0x3f3f3f3f
struct Query{
	int opt,u,v;
}Q[MAXN+5];
int fa[MAXN+5];
inline int Find(int x){return fa[x]==x?x:(fa[x]=Find(fa[x]));}
int main(){
	//freopen("tour.in","r",stdin);
	//freopen("tour.out","w",stdout);
	LL n=read(),m=read(),q=read();
	for(int i=0;i<n;i++)
		fa[i]=i;
	for(int t=1;t<=q;t++){
		int u=read(),v=read(),opt=read();
		Q[t]=(Query){opt,u,v};
	}
	int cnt=0,s=n;
	for(int t=1;t<=q;t++)
		if(Q[t].opt==0){
			int fu=Find(Q[t].u),fv=Find(Q[t].v);
			if(fu!=fv)
				fa[fu]=fv,s--;
		}
	for(int t=1;t<=q;t++)
		if(Q[t].opt==1){
			cnt++;
			int fu=Find(Q[t].u),fv=Find(Q[t].v);
			if(fu==fv){
				puts("No");
				return 0;
			}
		}
	if(cnt==0)
		puts((n-1<=m&&m<=n-s+1ll*s*(s-1)/2)?"Yes":"No");
	else puts((n<=m&&m<=n-s+1ll*s*(s-1)/2)?"Yes":"No");
    return 0;
}