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 FR first
#define SE second

using namespace std;

typedef long long ll;
typedef pair<int,int> pr;

namespace SETS {

int fa[100005],size[100005];

void init(int n) {
  for(int i=1;i<=n;i++) {
  	fa[i]=i;
  	size[i]=1;
  }
}

int find_father(int x) {
  return (fa[x]==x)?x:fa[x]=find_father(fa[x]);
}

void merge(int x,int y) {
  x=find_father(x);y=find_father(y);
  if (x==y) return;
  fa[x]=y;
  size[y]+=size[x];
}

}

pr e[100005];

int main() {
  int n,k;
  ll m;
  scanf("%d%d%d",&n,&m,&k);
  SETS::init(n);
  int cnt=0;
  for(int i=1;i<=k;i++) {
  	int x,y,z;
  	scanf("%d%d%d",&x,&y,&z);
  	x++;y++;
  	if (!z) SETS::merge(x,y);
  	else e[++cnt]=pr(x,y);
  }
  for(int i=1;i<=cnt;i++)
    if (SETS::find_father(e[i].FR)==SETS::find_father(e[i].SE)) {
    	puts("No");
    	return 0;
	}
  int s=n;
  for(int i=1;i<=n;i++)
    if (SETS::find_father(i)==i) s-=SETS::size[i]-1;
  if ((s<=2||m==n-1)&&cnt) {
  	puts("No");
  	return 0;
  }
  puts((((ll)s*(s-1)>>1)+n-s>=m)?"Yes":"No");
  return 0;
}