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;
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
typedef long long LL;
typedef unsigned long long uLL;
typedef pair<int,int> pii;

const int maxn=100111;

int n,q,a[maxn],b[maxn],c[maxn],f[maxn];
LL m;
int gf(int x){return x==f[x]?x:f[x]=gf(f[x]);}

int main()
{
	scanf("%d%lld%d",&n,&m,&q);
	for(int i=1;i<=q;i++)
	{
		scanf("%d%d%d",a+i,b+i,c+i);
		a[i]++;b[i]++;
	}
	if(m==n-1)
	{
		for(int i=1;i<=q;i++)if(c[i]==1)
		{
			puts("No");
			return 0;
		}
		puts("Yes");
		return 0;
	}
	for(int i=1;i<=n;i++)f[i]=i;
	for(int i=1;i<=q;i++)if(c[i]==0)f[gf(a[i])]=gf(b[i]);
	for(int i=1;i<=q;i++)if(c[i]==1&&gf(a[i])==gf(b[i]))
	{
		puts("No");
		return 0;
	}
	int cnt=0;
	for(int i=1;i<=n;i++)if(f[i]==i)cnt++;
	if(n+(LL)cnt*(cnt-3)/2<m)puts("No");
	else puts("Yes");
	return 0;
}