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<bits/stdc++.h>
#define ll long long
using namespace std;
struct aaa{
	ll x,y,op;
}a[200010];
ll n,m,M,ans,flag,fa[200010],mn,mx,sz;
ll find(ll x){
	if(x==fa[x])return x;
	return fa[x]=find(fa[x]);
}
int main(){
	ll x,y,i;
	scanf("%lld%lld%lld",&n,&M,&m);
	for(i=1;i<=n;i++)fa[i]=i;
	for(i=1;i<=m;i++){
		scanf("%lld%lld%lld",&a[i].x,&a[i].y,&a[i].op);a[i].x++;a[i].y++;
		if(a[i].op==0){
			x=find(a[i].x);y=find(a[i].y);
			if(x!=y)ans++,fa[x]=y;
		}
	}
	for(i=1;i<=m;i++)if(a[i].op==1){
		flag=1;if(find(a[i].x)==find(a[i].y))return puts("No"),0;
   }
   for(i=1;i<=n;i++)if(find(i)==i)sz++;
   if(flag&&sz<3)return puts("No"),0;
   mn=ans+sz;
   if(!flag)mn--;
   mx=ans+sz*(sz-1)/2;
   if(mn<=M&&M<=mx)puts("Yes");
    else puts("No");
}