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 <iostream>
#include <algorithm>
#include <cstring>
#define ll long long
using namespace std;
template <class T>
inline void rd(T &x)
{
	x=0; char c=getchar(); int f=1;
	while(!isdigit(c)){if(c=='-')f=-1; c=getchar();}
	while(isdigit(c)) x=x*10-'0'+c,c=getchar(); x*=f;
}
const int N=1e5+10;
int fa[N],n,q;
ll m;
struct Que { int x,y,c; }Q[N];
int find(int x) { return fa[x]==x?x:fa[x]=find(fa[x]); }
void FAIL() {
	printf("No");
	exit(0);
}
int main() {
	rd(n),rd(m),rd(q);
	for(int i=1;i<=n;++i) fa[i]=i;
	for(int i=1;i<=q;++i) {
		rd(Q[i].x),rd(Q[i].y),rd(Q[i].c);
		Q[i].x++,Q[i].y++;
		if(Q[i].c==0)
			fa[find(Q[i].x)]=find(Q[i].y);
	}
	int flg=0;
	for(int i=1;i<=q;++i) if(Q[i].c==1) {
		flg=1;
		if(find(Q[i].x)==find(Q[i].y)) FAIL();
	}
	int M=0;
	for(int i=1;i<=n;++i) M+=find(i)==i;
	if(m>M*(ll)(M-1)/2+(n-M)) FAIL();
	if(flg&&m<n) FAIL();
	printf("Yes");
	return 0;
}