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 <cstdio>
#include <cstring>
#include <algorithm>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)
#define rep(i,a) for(int i=lst[a];i;i=nxt[i])
using namespace std;

typedef long long ll;

int read() {
	char ch;
	for(ch=getchar();ch<'0'||ch>'9';ch=getchar());
	int x=ch-'0';
	for(ch=getchar();ch>='0'&&ch<='9';ch=getchar()) x=x*10+ch-'0';
	return x;
}

const int N=1e5+5;

int n,m,cnt,q,k,a[N],b[N],c[N],fa[N];

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

void No() {puts("No");exit(0);}

int main() {
	n=cnt=read();m=read();q=read();
	fo(i,1,q) a[i]=read()+1,b[i]=read()+1,c[i]=read();
	if (m==n-1) {
		fo(i,1,q) if (c[i]) No();
		puts("Yes");
		return 0;
	}
	fo(i,1,q) 
		if (!c[i]) {
			int x=get(a[i]),y=get(b[i]);
			if (x==y) continue;
			fa[y]=x;cnt--;k++;
		}
	fo(i,1,q) 
		if (c[i]) {
			int x=get(a[i]),y=get(b[i]);
			if (x==y) No();
		}
	if (cnt==2) {
		fo(i,1,q) if (c[i]) No();
		return 0;
	}
	if (m<k+cnt||m>k+(ll)cnt*(cnt-1)/2) No();
	puts("Yes");
	return 0;
}