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>
using namespace std;

typedef long long LL;

const int N=100000;

int n,cq,qx[N+9],qy[N+9],qv[N+9];
LL m;

void into(){
  scanf("%d%lld%d",&n,&m,&cq);
  for (int i=1;i<=cq;++i){
	scanf("%d%d%d",&qx[i],&qy[i],&qv[i]);
	++qx[i];++qy[i];
  }
}

int uni[N+9],vis[N+9];

int Query_uni(int k){return k==uni[k]?k:uni[k]=Query_uni(uni[k]);}

int ans;

void Get_ans(){
  if (m<n-1) return;
  for (int i=1;i<=n;++i) uni[i]=i;
  for (int i=1;i<=cq;++i){
	if (qv[i]) continue;
	int fx=Query_uni(qx[i]),fy=Query_uni(qy[i]);
	if (fx==fy) continue;
	uni[fx]=fy;
  }
  int flag=0;
  for (int i=1;i<=cq;++i){
	if (!qv[i]) continue;
	flag=1;
	if (Query_uni(qx[i])==Query_uni(qy[i])) return;
  }
  int cnt=0;
  for (int i=1;i<=n;++i){
	int fx=Query_uni(i);
	if (!vis[fx]) ++cnt,vis[fx]=1;
  }
  if (cnt==1||cnt==2) {ans=m==n-1&&!flag;return;}
  if (m<n&&flag) return;
  LL t=n-cnt+((LL)cnt*(cnt-1)>>1);
  ans=m<=t;
}

void work(){
  Get_ans();
}

void outo(){
  puts(ans?"Yes":"No");
}

int main(){
  into();
  work();
  outo();
  return 0;
}